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Projectile Motion is the motion experienced by an object when it gets projected into the air. The object is then subjected to acceleration under the influence of gravity. The path taken by the projectile or the object is a trajectory. The projectile motion formula is also known as the trajectory formula. The object moves along a curved route only.
The types of Projectile Motion Formula are:
- Horizontal Distance – x = Vx0t
- Horizontal Velocity – Vx = Vx0
- Vertical Distance, y – Vy0t – ½ gt2
- Vertical Velocity, Vy – Vy0 – gt
Key Terms: Projectile Motion, Acceleration, Velocity, Displacement, Range, Time of Flight, Maximum Height
Projectile Motion Formula
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The projectile motion formula can be represented as –
| Parameter | Projectile Motion Formula |
|---|---|
| Vertical Velocity | Vy = Vy0 – gt |
| Horizontal Velocity | Vx = Vx0 |
| Vertical Distance | y = Vy0t – ½ gt2 |
| Horizontal Distance | x = Vx0t |
Where,
- Vx refers to the velocity along x-axis
- Vxo refers to the initial velocity along x-axis
- Vy refers to the velocity along y-axis
- Vyo refers to the initial velocity along y-axis
- g refers to the acceleration due to gravity
- t refers to the time taken
The other important projectile motion formulas are as given below –
| Quantity | Value |
|---|---|
| Components of velocity at time t | vx = v0 cosθ0, vy = v0 sinθ0–gt |
| Position at time t | x = (v0 cosθ0)t, y = (v0 sinθ0)t – 1/2 gt2 |
| Equation of path of projectile motion | y = (tan θ0)x – gx2/2(v0cosθ0)2 |
| Time of maximum height | tm = v0 sinθ0 /g |
| Time of flight | 2tm = 2(v0 sinθ0/g) |
| Maximum height of projectile | hm = (v0 sinθ0)2/2g |
| Horizontal range of projectile | R = v02 sin 2θ0/g |
| Maximum horizontal range ( θ0 = 45° ) | Rm = v02/g |
Projectile Motion Formula – Solved ExampleExample: Aditya is on the top of the building while Jack is standing down. If Aditya tosses a ball with a 30m/s velocity at the angle of 70° then at 3 sec what height will the ball reach? Solution: Given: Vyo = 30 m/s, Δ t = 3s The vertical velocity in the y-direction is expressed as Vy = Vyo sin 70° Vy = 30 x sin 70 Vy = 23.22 m/s Therefore, at 3 seconds the ball will reach 23.22 m/s. |
Also Read:
Projectile Motion
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Projectile motion is applied to an object or particle (a projectile) when it’s projected near the Earth's surface. The lateral and vertical motions in projectile motion are independent of one another; that is, neither motion impacts the other.
Projectile Motion
- On Earth, the amplitude and direction of acceleration change with altitude and latitude/longitude.
- On a small scale, this results in an elliptic trajectory that is remarkably similar to a parabola.
- If an object was thrown and the Earth was replaced with a black hole of similar mass, it would be clear that the ballistic trajectory would be part of an elliptic orbit around the black hole, not a parabola that continues to infinity.
- Unless modified by other objects such as the Moon or the Sun, the trajectory can be circular, parabolic, or hyperbolic at greater speeds.
Also Read: Law of Inertia
Time of Flight
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A projectile motion's time of flight is exactly what it sounds like. Projectile Motion Formula of Time of Flight is the period of time between when an object is projected and when it reaches the surface. The duration of the flight is determined by the object's initial velocity and the projection angle.
Time of Flight, t = 2 × \(Vy \over g\)
| Example: An object is projected with a velocity of 20 ms−1 at 50° to the horizontal plane. Determine the time of flight of the projectile. Solution: Initial Velocity u = 20 ms−1 |
Range of Projectile
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The horizontal displacement of the projectile determines its range. Gravity only acts vertically; hence there is no acceleration in this direction. shows the range line. The range of the projectile, like its time of flight and maximum height, is a function of its initial speed.
Range of projectile
| Range of The Projectile: R = 2 × Vx × Vy / g |
Maximum Height Range
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When the projectile approaches zero vertical velocity, it reaches its maximum height. The vertical component of the vector field will point down from such a location. The range of the projectile is the horizontal displacement of the projectile and is determined by the object's starting velocity. The projectile motion formula for maximum height range is given as –
Maximum Height: hmax = (v02) / 2g
Maximum Height
Here:
- V = Velocity
- Vx = Horizontal Velocity
- Vy = Vertical Velocity
- a = Angle of Launch
- h = Initial Height
- t = Time of Flight
- d = Distance (range)
- hmax = Maximum Height
| Example: A man wants to throw an object into a window located on the second floor which is 12 m from the ground. Determine its speed at an angle of 30 degrees. Solution: After applying the respective formula for maximum height of a projectile, [S = (u sinθ) 2 / 2g] |
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Acceleration, Velocity and Displacement
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A projectile has a vertical acceleration of 9.8 m/s, a downward acceleration of 0 m/s, and no horizontal acceleration. This means the vertical velocity changes by 9.8 m/s every second while the horizontal velocity remains constant.
- Since there is only acceleration in the vertical plane, the velocity in the horizontal plane is static, equal to v(0)cos.
- The vertical velocity of the projectile is similar to that of a particle in free fall.
- In this instance, the acceleration is constant and equal to g. The components of acceleration of projectile motion formula are as follows:
| ax = 0, ay = -g |
- A projectile's horizontal velocity is fixed (it never changes in value), Gravity causes a vertical acceleration of 9.8 m/s, which is downward.
- Each second, a projectile's vertical velocity changes by 9.8 meters per second.
- A projectile's horizontal motion is separate from its vertical motion.
- The range of the projectile is the horizontal displacement of the projectile and is determined by the object's starting velocity.
- The range of a projectile will be the same if it is projected at the same initial speed but at two complementary angles of projection.
- At any time (t ), the projectile's displacements(x and y) in horizontal and vertical planes are given by:
| x = v0t cos(θ), y = v0t sin(θ) − ½ gt2 |
Read More: Speed Distance Time Formula
Things to Remember
- Projectile motion occurs when an object is projected near the Earth's surface and moves along a curved route only under the influence of gravity.
- Time of Flight is the period of time between when an object is projected and when it reaches the surface.
- A projectile has a vertical acceleration of 9.8 m/s, a downward acceleration of 0 m/s, and no horizontal acceleration.
- The range of the projectile is the horizontal displacement of the projectile and is determined by the object's starting velocity.
- Projectile Motion Formula –
- Horizontal Velocity, Vx = Vx0
- Horizontal Distance, x = Vx0t
- Vertical Velocity, Vy – Vy0 – gt
- Vertical Distance, y, Vy0t – ½ gt2
Previous Year Questions
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Sample Questions
Ques. An item is released at 20 m/s in a trajectory that makes a 25° upward angle with the horizontal. [5 marks]
(a) What is the highest height that the object can reach?
(b) What is the object's entire flight time (from launch to contact with the ground)?
(c) What is the horizontal range (maximum x above ground) of the object?
(d) What is the magnitude of the velocity of the object just before it hits the ground?
Ans.
- The formulas for the components Vx and Vy of the velocity and components x and y of the displacement are given by
Vx = cos(θ)y = sin(θ) - gt
x = cos(θ) ty = sin(θ) t - (1/2) gt2
In the problem = 20 m/s, θ = 25° and g = 9.8 m/s2
The height of the projectile is given by the component y, and it reaches its maximum value when the component y is equal to zero. It's when the projectile changes from moving upward to moving downward.
Vy = V0 sin(θ) - gt = 0
Now, solving for t, we get,
t = V0 sin(θ) / g = 20 sin(25°) / 9.8 = 0.86 seconds
We can find the maximum height by substituting t by 0.86 seconds in the formula for y,
Maximum Height, y(0.86) = 20 sin(25°)(0.86) - (1/2) (9.8) (0.86)2 = 3.64 meters
- The time of flight is the interval of time between when the projectile is launched: t1 and when the projectile touches the ground: t2. At t = t1 and t = t2, y = 0 (ground). Hence
V0 sin(θ) t - (½) gt2 = 0
Now, after solving for t, we get,
t (V0 sin(θ) - (½) gt) = 0
We will obtain two equations likewise,
t = t1 = 0 and t = t2 = 2 V0 sin(θ) / g
Time of flight = t2 - t1 = 2 (20) sin(θ) / g = 1.72 seconds.
- The time of flight, t2 = 2 V0 sin(θ) / g
The horizontal range is the horizontal distance given by x at t = t2
Thus, range = x(t2) = V0 cos(θ) t2 = 2V0 cos(θ) V0 sin(θ) / g
= V02 sin(2θ) / g
= 202 sin (2(25°)) / 9.8
= 31.26 meters
- An object hits the ground at, t = t2 = 2 V0 sin(θ) / g . . . (Determined in part ‘b’ above)
The components of velocity at t are given by,
Vx = V0 cos(θ) Vy = V0sin(θ) – gt
The components of the velocity at t = 2 V0 sin(θ) / g are given by
Vx = V0 cos(θ)
= 20 cos(25°) Vy
= V0 sin (25°) - g (2V0 sin (25°) / g)
= -V0 sin(25°)
The magnitude V of the velocity can be given by,
V = √[Vx2 + Vy2]
= √[(20 cos(25°))2 + (-V0 sin(25°))2]
= 20 m/s
Ques. The projectile is launched at an inclination of 22° with an initial velocity of 15 m/s up an incline plane that makes a 10° angle with the horizontal from point O. At point M, the projectile collides with the inclination plane. [2 marks]
(a) Calculate how long it takes the projectile to strike the inclination plane.
(b) Find the distance OM.
Ans.
(a) The x and y components of the displacement are given by
x = V0 cos(θ) t y = V0 sin(θ) t - (1/2) gt2
with θ = 22 + 10 = 32° and V0 = 15 m/s
The relationship between the coordinate x and y on the incline is given by
tan(10°) = y / x
Substitute x and y by their expressions above to obtain
tan(10°) = ( V0 sin(θ) t - (1/2) g t2) / V0 cos(θ) t
Simplify to obtain the equation in t
(1/2) gt + V0 cos(θ) tan(10°) - V0 sin(θ) = 0
t = 1.16 s
(b) OM = √[ (V0 cos(θ) t)2 + ( V0 sin(θ) t - (1/2) gt2 )2 ]
OM (t = 1.16) = √[ (15 cos (32) 1.16)2 + (15 sin (32) 1.16 - (½) 9.8 (1.16)2)2]
= 15 meters
Ques. Two balls, A and B, each weighing 100 grams and 300 grams respectively, are pushed horizontally from a 3 meter high table. While ball A is pushed with a 10 m/s initial velocity, ball B is propelled with a 15 m/s initial velocity. [2 marks]
(a) Find the time it takes each ball to hit the ground.
(b) What is the difference in the distance between the points of impact of the two balls on the ground?
Ans. The two balls are subject to the same gravitational acceleration and therefore will hit the ground at the same time t found by solving the equation
- -3 = -(1/2) gt2
t = √ (3(2)/9.8) = 0.78 s
- Horizontal distance XA of ball A
XA = 10 m/s * 0.78 s = 7.8 m
Horizontal distance XB of ball B
XB = 15 m/s * 0.78 s = 11.7 m
Difference in distance XA and XB is given by
|XB - XA| = |11.7 - 7.8|
= 3.9 m
Ques. What are the 3 types of projectile motion? [2 marks]
Ans. The three types of Projectile Motions are,
- Oblique projectile motion.
- Horizontal projectile motion.
- Projectile motion on an inclined plane.
Ques. What causes projectile motion in objects? [1 mark]
Ans. A projectile is an object on which the only force acting is gravity. Vertical acceleration is caused by gravity's influence on the projectile's vertical motion. The tendency for any moving object to maintain a constant velocity causes the projectile to move horizontally.
Ques. Highlight the point of trajectory at which the projectile motion has the minimum speed? [1 mark]
Ans. A projectile motion has minimum speed at the highest point of its trajectory. It has minimum speed because the horizontal speed of the projectile usually remains constant during the motion. Ideally, the projectile records minimum speed because its vertical components attain zero when at the maximum height.
Ques. An object is projected horizontally with a velocity of 10 ms, which makes an angle about 45 degrees. With the given equation, determine the time of flight. [3 marks]
Ans. The object is projected horizontally and has a velocity = 10 ms
The angle of projection = 45 degrees
Therefore, the Time of Flight, as per formula, can be given as, T = (2u sinα) / g
Wherein,
u = initial projectile velocity
α = angle of projection
g = acceleration due to gravity
Thus, it can be said,
Time of flight T = ( 2 × 10 × sin 45 ) / 9.8 = 1.44 s
Ques. Highlight the lowest speed of an object during its projectile when fired at an angle of 30 degrees, with a speed of 5 m/s. [3 marks]
Ans. The object fired at an angle of = 30 degrees
The speed of the object being fired = 5 m/s
Now, we are aware that the lowest speed of a projectile motion can be witnessed when it reaches maximum flight.
It is clear that the vertical component of velocity is zero at maximum height, and the object only has a horizontal component.
Therefore, the lowest speed of the object can be given by, (5 cos 30) m/s, which is, 5 × \(\frac{\sqrt{3}}{2}\) = 4.33 m/s.
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