Question

A player throws a ball vertically upwards with a
velocity of 29.4 ms', then
(i) What is the direction of acceleration acting on
the ball?
(ii) What is the velocity and acceleration of the
ball at the highest point?
(iii) To what height the ball rises and after how
much time does it return back into the hands of
the player? (g = 9.8 ms ?).​

Answer 1

Answer:-

h = 44.1 m

T = 4s

Given :-

$u = 29.4 m/s\\ g = -10 m/s^2$

To find :-

(ii) What is the velocity and acceleration of the

ball at the highest point?

(iii) To what height the ball rises and after how

much time does it return back into the hands of

the player?

Solution:-

1) When ball is thrown upward it velocity is retarded and at highest point it becomes 0 ; V = 0 m/s.

• The acceleration of the body at maximum height will be equal to g I. e, = 9.8 m/s².

The height attained by ball is :-

→$2gh = v^2 - u^2$

→$2 \times 9.8 \times h = (0) ^2-(29.4) ^2$

→$-19.6h = - 864.36$

→$h = \dfrac{-864.36}{19.6}$

→$h = 44.1 m$

Hence,

maximum height attained by body is 44.1 m.

Time of flight is given by :-

→$T = \dfrac{2u}{g}$

→$T = \dfrac{2 \times 19.6}{9.8}$

→$T = \dfrac{39.2}{9.8}$

→$T = 4 s$

hence,

Time taken by ball to return in hand of player is 4s.