Question

13. A ball is thrown vertically upwards. It returns
6 s later. Calculate : (i) the greatest height reached
by the ball, and (ii) the initial velocity of the ball.
(Take g = 10 m s-2)
​

Answer 1

Answer:

ANSWER

The ball is thrown up and it returns in 6sec,

Time of accent = Time of decent

So, time to reach the highest point =6/2=3s

By 2nd equation of motion

s=ut+

2

1

gt

2

To calculate height, consider motion of the ball from highest point to the ground

H=

2

1

gt

2

=

2

1

×10×3

2

=45 m

To calculate projection velocity, consider motion of the ball from projection to return to the point of projection.

0=ut−

2

1

gt

2

u=10×6/2=30 m/s

Answer 2

Greatest height is 180m

S=ut+1/2at^2

Where
u=0
t=6s
a=10m/s^2

So

S=1/2at^2
S=1/2*10*36

When we solve get answer as 108m

Initial velocity = 0