Question

A golfer hit a golf-ball from the top of a building 25 ft off the ground with an initial velocity of 45 feet per second. What is the maximum height that the golf ball will reach?

Answer 1

56.641 feet

Explanation:

Use the equation
`h(t)=-16t^2+v_0t+h_0` to find the vertical distance of an object travelling at a speed of
`v_0\text{ft}/\text{s}` at initial height
`h_0\text{ft}` after
`t` seconds.

First, we find the time at which the ball will reach maximum height:

`t=-(b)/(2a)\\\\t=-(45)/(2(-16))\\ \\t=-(45)/(-32)\\\\t=(45)/(32)\approx1.406`

Second, we find the vertical distance of the ball after
`t=(45)/(32)` seconds:

`h((45)/(32))=-16((45)/(32))^2+45((45)/(32))+25=56.641`

Therefore, the golf ball will reach a maximum height of 56.641 feet after 1.406 seconds.

Answer 2

The maximum height is 180 feet

Explanation: