Projectile Motion is an important topic in the Physics section in TS EAMCET exam. Practising this topic will increase your score overall and make your conceptual grip on TS EAMCET exam stronger.
This article gives you a full set of TS EAMCET PYQs for Projectile Motion with explanations for effective preparation. Practice of TS EAMCET Physics PYQs including Projectile Motion questions regularly will improve accuracy, speed, and confidence in the TS EAMCET 2026 exam.
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TS EAMCET PYQs for Projectile Motion with Solutions
1.
A gill throws a stone horizontally from the roof of a house 12 in above the ground with a speed of If 15 m/s. Neglecting air resistance, the time it takes tor the stone to reach the ground is- 1.55 sec
- 3.1 sec
- 2.34 sec
- 4.10 sec
2.
A cannon ball is fired from a building of height h at the sea shore with a velocity v at an angle of a upwards from the horizontal, targeting a ship at a horizontal distance of S from the shore, as shown in the figure. If 'g' denotes the acceleration due to gravity, tan $\alpha$ governed by the quadratic equation is- $\frac{g}{2v^{2}} \tan^{2} \alpha+\frac{h}{s} \tan\alpha + \frac{gs }{2v^{2}}=0$
- $\frac{g}{2v^{2}} \tan^{2} \alpha - \frac{h}{s} \tan\alpha + \frac{gs }{2v^{2}}=0$
- $\frac{g}{2v^{2}} \tan^{2} \alpha + \tan\alpha + \frac{gs }{2v^{2}} - \frac{h}{s}=0$
- $\frac{g}{2v^{2}} \tan^{2} \alpha - \tan\alpha + \frac{gs }{2v^{2}} - \frac{h}{s}=0$
3.
A ball is projected vertically up with speed \( V_0 \) from a certain height \( H \). When the ball reaches the ground, the speed is \( 3V_0 \). The time taken by the ball to reach the ground and height \( H \) respectively are:- \( \frac{V_0}{g} \), \( \frac{V_0^2}{2g} \)
- \( \frac{V_0}{g} \), \( \frac{3V_0^2}{2g} \)
- \( \frac{2V_0}{g} \), \( \frac{V_0^2}{2g} \)
- \( \frac{3V_0}{g} \), \( \frac{3V_0^2}{2g} \)
4.
The maximum range of a projectile is 80 m. If the projectile is projected with the same speed at an angle of \( \frac{\pi}{12} \) with the horizontal, then the range of the projectile is:- \(\mathbf{40 \text{ m}}\)
- \(80 \text{ m}\)
- \(20 \text{ m}\)
\(60 \text{ m}\)
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