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NEET PHYSICSMOTION IN A PLANEEasy

Question

A cricket ball is thrown by a player at a speed of 20 m/s20 \text{ m/s} in a direction 3030^{\circ} above the horizontal. The maximum height attained by the ball during its motion is: (Take g=10 m/s2g=10 \text{ m/s}^2)

A

5 m5 \text{ m}

B

10 m10 \text{ m}

C

20 m20 \text{ m}

D

25 m25 \text{ m}

Step-by-Step Solution

The maximum height hmh_m reached by a projectile is given by the formula: hm=(v0sinθ0)22gh_m = \frac{(v_0 \sin \theta_0)^2}{2g} where v0v_0 is the initial speed, θ0\theta_0 is the angle of projection, and gg is the acceleration due to gravity .

Given: Initial speed, v0=20 m/sv_0 = 20 \text{ m/s} Angle of projection, θ0=30\theta_0 = 30^{\circ}

  • Acceleration due to gravity, g=10 m/s2g = 10 \text{ m/s}^2

Substituting these values into the equation: hm=(20sin30)22×10h_m = \frac{(20 \sin 30^{\circ})^2}{2 \times 10} hm=(20×0.5)220h_m = \frac{(20 \times 0.5)^2}{20} hm=(10)220h_m = \frac{(10)^2}{20} hm=10020=5 mh_m = \frac{100}{20} = 5 \text{ m}

Thus, the maximum height attained by the ball is 5 m5 \text{ m}.

Exam Context & Concepts Covered

This question aligns with the NEET PHYSICS syllabus, specifically targeting concepts from MOTION IN A PLANE. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

PHYSICSMOTION IN A PLANEcricketthrownplayerdirectionhorizontal

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