Answer:
a.) Maximum height = 127.4 metre
b.) Time = 5.1 seconds
Step-by-step 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