Question

A ball is thrown at 32 m/s [up] from the edge of a cliff 45 m high. What is the maximum height of the ball and at what speed will the ball strike the ground?

Answer 1

`h_("max") ~~97.245` m

`v_("ground") ~~ -43.656 **bolded text**` m/s

Explanation:

Given: A ball is thrown at 32 m/s up on a cliff 45 m high. What is maximum height and what is the speed when the ball strikes the ground?

The equation for the height : `h(t) = 1/2 g t^2 + v_o t + h_o`

where `g = -9.8 m/s^2; " "v_o = 32; " "h_o = 45`

For the given data:

`h(t) = -4.9t^2 + 32t + 45`

`h_("max")` happens with the velocity `v(t) = 0`. Find the time it takes for the velocity to be = 0 :

`v(t) = h'(t) = -9.8 t + 32 = 0`

`9.8 t = 32; " " t = 32/9.8 ~~ 3.2653` sec

Find the maximum height:

`h_("max") = h(3.2653) = -4.9 (3.2653)^2 + 32(3.2653) + 45`

`h_("max") ~~ 97.245` m

Find the time it takes for the ball to hit the ground (h = 0) :

`0 = -4.9t^2 + 32t + 45`

Use the quadratic formula:

`t = (-32 +- sqrt(32^2 - 4(-4.9)(45)))/(2(-4.9))`

We are only interested in the positive answer:

`t = (-32 - sqrt(1906))/(2(-4.9)) ~~7.72` sec

Find the speed when the ball hits the ground :

`v(7.72) = -9.8 (7.72) + 32 ~~-43.656` m/s