Answer:
100 ft
16 ft/s
16 ft/s
Explanation:
t = Time taken
u = Initial velocity
v = Final velocity
s = Displacement
a = Acceleration due to gravity = 32 ft/s² (negative upwards and positive downwards)
The function here is of the form which is one of the equations of motion
[tex]s=ut+\frac{1}{2}at^2\\\Rightarrow s=80t+\frac{1}{2}32t^2\\\Rightarrow s=80t+16t^2[/tex]
From another equation of motion
[tex]v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{0^2-80^2}{2\times -32}\\\Rightarrow s=100\ ft[/tex]
The maximum height the ball will reach is 100 ft
When s = 96 ft
[tex]v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times -32\times 96+80^2}\\\Rightarrow v=16\ ft/s[/tex]
The velocity of the ball when it is 96 ft above the ground on its way up is 16 ft/s
On the way down when it will be 96 ft above the ground it would travel 100-96 = 4 ft from the maximum height
[tex]v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 32\times 4+0^2}\\\Rightarrow v=16\ ft/s[/tex]
The velocity of the ball on the way down when it is 96 ft above the ground is 16 ft/s
