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

Answer:

The maximum height is 180 feet

Step-by-step explanation:

Answer 2

Answer:

56.641 feet

Step-by-step 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:

$t=-\frac{b}{2a}\\\\t=-\frac{45}{2(-16)}\\ \\t=-\frac{45}{-32}\\\\t=\frac{45}{32}\approx1.406$

Second, we find the vertical distance of the ball after $t=\frac{45}{32}$ seconds:

$h(\frac{45}{32})=-16(\frac{45}{32})^2+45(\frac{45}{32})+25=56.641$

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