Question

A ball is thrown upwards with initial velocity of 72 km per hour. Calculate the

i. maximum height attained by the ball

ii. velocity which wich it reaches the ground

iii. time taken to reach the ground​

Answer 1

Answer:

Explanation:

1) max height =$u^{2}$/2g

now once you know h

2)velocity when it reaches the ground =$\sqrt{2gh}$

3) time taken=2u/g

hope this helps

Answer 2

Answer:

Step by step explanation:-

Given initial velocity, u = 72 km/hr = 72×1000m DIVIDE 3600 s = 20 m/s

To calculate the time taken by to return initial position, use equation of first motionv = u +gtHere, v is initial velocity u is final velocity t is time taken and g is acceleration due to gravity and its value is 9.8m/S (square)

To return the initial position, the final velocity v =0.

Substitute the given value, we get

0 = 20 m/s -(9.8m/S (square) t

t = 20 m/s DIVIDE 9.8 m/s (square) = 2.04/s

So, time taken by ball to return its initial position= 2t = 2(2s) = 4 s

To calculate the maximum height, use third equation of motion

V(square) = U(square) +2gh

Here, h is maximum height.At maximum height the final velocity is zero.

At maximum height the final velocity is zero.

Substitute the value, we get

0=(20m/s) square -- 2(9.8m/S (square)) h

h= 400m/s DIVIDE 19.6 = 20.4m

Thus, time taken to return initial position is 4 s and the maximum height is 20 m.