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

more than, less

less than, more

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