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Standard XII

Physics

1st Equation of Motion

A ball is thr...

Question

A ball is thrown with a velocity of $5 m / s$ towards the ground from the top of a building of height $10 m$ . What is the change in its speed, when the ball is about to hit the ground? Take $g = 10 m / s^{2}$ .

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Solution

We have been given:
$u = - 5 m / s$
$h = - 10 m$
$a = - 10 m / s^{2}$

Using 3rd equation of kinematics:
$v^{2} = u^{2} + 2 a s$
$v^{2} = (- 5)^{2} + 2 (- 10) (- 10)$
$v^{2} = 25 + 200 = 225$
$v = \sqrt{225} = \pm 15 m / s$
As the ball is travelling downwards, the final velocity will also be in downward direction (negative).
Hence, $v = - 15 m / s$

So, the change in speed of the ball:
$| v | - | u | = 15 - 5 = 10 m / s$

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1st Equation of Motion

Standard XII Physics

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A ball is thrown with a velocity of 5 m/s towards the ground from the top of a building of height 10 m. What is the change in its speed, when the ball is about to hit the ground? Take g=10 m/s2.

A ball is thrown with a velocity of $5 m / s$ towards the ground from the top of a building of height $10 m$ . What is the change in its speed, when the ball is about to hit the ground? Take $g = 10 m / s^{2}$ .