Question

A ball is thrown straight up, with no air resistance. Its speed
at *half* the maximum height that it can reach is 18.0 m/s.
Calculate the maximum height it reaches, in meters. Use *g*
= 10 m/s^{2}.

Answer #1

Using 3rd kinematic equation, when ball reaches max height of 'H', then

V^2 = U^2 + 2*a*H

U = Initial upward speed

V = final speed at max height = 0 m/s

a = acceleration due to gravity = -g = -10 m/s^2

So,

0^2 = U^2 - 2*g*H

U^2 = 2*g*H

U = sqrt (2*g*H)

Now again use 3rd kinematic equation for 'H/2' height

V1^2 = U^2 + 2*a*(H/2)

V1 = speed at 'H/2' height = 18.0 m/s

So,

18.0^2 = 2*g*H + 2*(-g)*(H/2)

g*H = 18.0^2

H = 18.0^2/10

**H = 32.4 m = max height reached by the ball**

**Let me know if you've any query.**

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