Chapter 3 Laws Of Motion | Exercise 3E

Question

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= $9.8m/s^2$ )

Accepted 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

Chapter 3 Laws Of Motion | Exercise 3E

Chapters in Selina Solutions CONCISE subject class 9-ICSE-Physics

Exercises in Chapter 3 Laws Of Motion

Question 44