Question

a ball is thrown vertically upward with a velocity 20 m/s from the top of a multi-storey building. the height of the point from where the ball is thrown is 25 m from the ground, i) how high the ball rise? ii) with what velocity strike the ground?

Answer 1

i) you know the initial velocity, final velocity, acceleration, and you are trying to work out the displacement, so use this formula:

v^2 = u^2 + 2as

Where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the displacement.

Rearrange for s:

2as = v^2 - u^2
s = (v^2 - u^2)/2a

a is the acceleration due to gravity (g = 9.8 m/s^2 at sea level to 2.s.f), u is the initial velocity of 20 m/a, v will be zero as the ball reaches maximum height when it comes to a stop midair.

Sub in values:

s = -(20)^2 / 2•-9.8
s = -400/-19.6
s = 20.4 m (1.d.p)

Maximum height is the value we just worked out plus the initial height of 25m, therefore:

Max height = 20.4 + 25 = 45.4 m

ii) we know the initial velocity, the initial displacement, the acceleration, and we want to work out final velocity. So we can use the same formula:

v^2 = u^2 + 2as

v = sqrt(u^2 + 2as)

Take the downward direction as positive:

u = -20 m/s a = 9.8 m/s^2 s = 25 m

v = sqrt(400 + 490)
v = sqrt(890)

v = 30 m/s (2.s.f)