Physics - Kinematics Question with Solution | TestHub

PhysicsKinematics2-D & 3DHard2 minQB

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Passage / Comprehension

A ball is thrown from the top of a 75 m high tower with velocity 20 m/s at 30° dip angle. Then

What will be the velocity with which it hits the ground :-

Answer:

Solution:

By using $s = u t + \frac{1}{2} a t^{2} \Rightarrow - 75 = - 10 T + \frac{1}{2} (- 10) T^{2}$

$\Rightarrow T^{2} + 2 T - 15 = 0$

$\Rightarrow T^{2} + 5 T - 3 T - 15 = 0$

$\Rightarrow T (T + 5) - 3 (T + 5) = 0 \Rightarrow (T + 5) (T - 3) = 0 \Rightarrow T = 3 s$

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Stream:JEESubject:PhysicsTopic:KinematicsSubtopic:2-D & 3D

⏱ 2mℹ️ Source: QB

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