A ball is projected at an initial speed of 20.0 meter per second making an angle of 45.0 with horizontal. What is the maximum height it will reach?

10.2 metres

Explanation:

Given that a ball is projected at an initial speed of 20.0 meter per second making an angle of 45.0 with horizontal. What is the maximum height it will reach?

Solution

To get the maximum height, let us use the formula

V^2 = U^2 sin^2ø - 2gH

At maximum height V = 0

U^2 sin^2ø = 2gH

Substitute all the parameters into the formula

20^2 ( sin 45 )^2 = 2 × 9.8 × H

400 × 0.5 = 19.6 H

Make H the subject of formula

H = 200 / 19.6

H = 10.204 metres.

Therefore, the maximum height reached by the projected ball is 10.2 metres.

can be answered using momentum.

momentum = p

p1= mass1 * velocity1

p1 = p2

therefore

p1 = 6kg * 4m/s

p1 = 24kgm/s

p2=m2 * v2

24 kgm/s = 7kg * v2

v2 = 24 kgm/s / 7kg

v2= 3.42 m/s