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Selina Solutions Class 9 Physics Solutions for Exercise 3E in Chapter 3 - Chapter 3 Laws Of Motion
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A ball is thrown vertically upwards with an initial velocity of 49m/s. Calculate: (i) the maximum height attained, (ii) the time taken by it before it reaches the ground again. (Take g=9.8m/s^2)
Answer:
Given: initial velocity u=49m/s, g=9.8m/s^2
(i) Assume ‘h’ to be the height, velocity = 0m/s at the greatest height
We know from the equation of motion,
\begin{array}{l} v^{2}-u^{2}=2 g h \\ \Rightarrow 0-(49)^{2}=2(-9.8)(h) \\ \Rightarrow h=(49)^{2} / 19.6 \\ \Rightarrow h=122.5 m \end{array}
(ii) Time taken before the ball reaches the ground is given by t=2u/g
t=2(49)/9.8
= 10s
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