Question

A ball is thrown upwards from a rooftop, 80 m above the ground. It will reach a maximum vertical height and then fall back to the ground. The height of the ball from the ground at time t is h, and is given by h=−16t2+64t+80h=−16t2+64t+80.   Find the time it will take before hitting the ground.  Consider the following HINTS:

1. When the ball hits the ground, the height is zero.

2. Use the concept of solving quadratic equations to solve this problem.

3. The time is "positive".​

Answer 1

Answer:

a. We use the equation provided, plugging in t=1. We get h=-16+64+80=128.

b. Whenever you see the word "maximum" (or "minimum," in a different setting) think "set derivative equal to zero/undefined." To determine the maximum height, we need to derivative of the height equation:

h' = -32t + 64

Set h'=0 and solve for t:

0=-32t+64

t=2

Thus, maximum height occurs at t=2. Plug this in to the original equation for h:

h=-16*(2)^2+64*2+80=144

(This answer looks good because it's bigger than our answer for part a! Having a basic understanding of what the ball is doing can save you from making silly mistakes.)

c. The ground is height h=0. So, we solve for t:

0=-16t^2+64t+80

We use the quadratic formula to find t=-1 and t=5. But, t=-1 doesn't make any sense (-1 second?) so our solution is t=5.

Answer 2

Given: A ball is thrown upwards from a rooftop, 80 m above the ground. It will reach a maximum vertical height and then fall back to the ground. The height of the ball from the ground at time t is h, and is given by h=−16t2+64t+80.

To find: The time it will take before hitting the ground.

Solution:

According to the question, the following equation can be used to find the time taken for the ball to reach the ground.

$H = -16t^{2} + 64t +80$

Here, H is the final height which is zero because when the ball touches the ground, the ball is 0 meters from the ground. Besides, t is the time taken for it to reach the ground. Now, the equation can be written as follows.

$0 = -16t^{2} + 64t + 80$

$16t^{2} - 64t - 80 = 0$

$t^{2} - 4t - 5 = 0$

Now, the quadratic equation formed gives the value of t.

$t = 5$ or $t = -1$

Since the time cannot be negative, the time taken is 5 seconds.

Therefore, the time it will take before hitting the ground is 5 seconds.