A ball is thrown straight up with an initial velocity of 50 m/s.

a) How high will it go?

b) How long does it take to reach this height?

a.) Maximum height = 127.4 metre

b.) Time = 5.1 seconds

Explanation:

Given that initial velocity U = 50m/s

a.) How high will it go means maximum height.

At maximum height, final velocity V = 0

Using the equation 3 of linear motion

V^2 = U^2 - 2gH

As the ball is going up, g will be negative.

0 = 50^2 - 2 * 9.81H

2500 = 19.62H

H = 2500/19.62

H = 127.4 meters

b.) The time taken to reach the maximum height

Using the 2nd equation of motion

h = Ut + 1/2gt^2

127.4 = 50t - 1/2 * 9.8t^2

127.4 = 50t - 4.9t^2

4.9t^2 - 50t + 127.4 = 0

This will lead to quadratic equation

Where a = 4.9, b = - 50, c = 127.4

Using quadratic formula will lead to

50 + / - root (50^2 - 4 * 4.9 * 127.4) / 2 * 4.9

50 + / - root (2500 - 2497.04) / 9.8

50+ / - root (2.96) / 9.8

50+ / - 1.72/9.8

(50 + 1.72) / 9.8 or (50 - 1.72) / 9.8

51.72/9.8 = 5.28s

Or 48.28/9.8 = 4.9s

To test for this let's assume it's a perfect parabola. Using vertex formula,

t = - b/2a

t = 50/9.8

t = 5.1 m/s