Question

A ball is thrown vertically upward with an initial speed of 72km/hr. Calculate the time taken to return to the thrower and the maximum height reached

Answer 1

Answer:

solved

Explanation:

72km/hr = 20 m/s = u

t = 2u/g = 40/10 = 4 s

Calculate the time taken to return to the thrower = 4s

maximum height reached = u²/2g = 400/20 = 20m.

Answer 2

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

Explanation:

Given initial velocity, u = 72 km/hr = $\frac{72\times1000 m}{3600s} = 20 m/s$.

To calculate the time taken by to return initial position, use equation of first motion

v = u +gt

Here, v is initial velocity u is final velocity t is time taken and g is acceleration due to gravity and its value is 9.8$m/s^{2}$.

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

Substitute the given value, we get

$0 = 20 m/s - (9.8 m/s^2)t$

$t =\frac{20 m/s}{9.8 m/s^{2} } =2.04 s$.

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

$v^{2} =u^{2} +2gh$

Here, h is maximum height.

At maximum height the final velocity is zero.

Substitute the value, we get

$0=(20m/s)^2-2(9.8m/s^2)h$

$h=\frac{400m/s}{19.6} =20.4 m$

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

Learn more: equation of motions

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