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NCERT Solutions Class 9 Science Solutions for Exercises in Chapter 8 - Motion

Question 6 Exercises

A ball is gently dropped from a height of 20 m. If its velocity increases uniformly at the rate of 10 ms^{-2} , with what velocity will it strike the ground? After what time will it strike the ground?

Answer:

Given, the initial velocity of the ball (u) = 0 (since it began at the rest position)

Distance travelled by the ball (s) = 20m

Acceleration (a) = 10 ms^{-2}

As per the third motion equation,

2 a s=v^{2}-u^{2}

Therefore, v^{2}=2 a s+u^{2}

\begin{aligned} &=2^{*}\left(10 \mathrm{ms}^{-2}\right)^{*}(20 \mathrm{m})+0\\ &v^{2}=400 m^{2} s^{-2}\\ &\text { Therefore, } v=20 \mathrm{ms}^{-1} \end{aligned}

The ball hits the ground with a velocity of 20 meters per second.

As per the first motion equation, v=u+a t

\begin{array}{l} \text { Therefore, } t=\frac{v-u}{a} \\ =\frac{(20-0) m s^{-1}}{10 m s^{-2}} \end{array}

= 2 seconds

Therefore, the ball reaches the ground after 2 seconds.

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