Question

A bullet is fired straight up from a BB gun with initial velocity 1,120 feet per second at an initial height of 8 feet. Use the formula h=−16t2+v0t+8 to determine how many seconds it will take for the bullet to hit the ground. That is, when will h=0

Answer 1

t≈70 seconds

Explanation:

h=−16t2+v0t+8

We know the velocity, v0, is 1,120 feet per second.

The height is 0 feet. Substitute the values.

0=−16t^2+1,120t+8

Identify the values of a, b, and c.

a=−16,b=1,120,c=8

Then, substitute in the values of a, b, and c.

t=−(1,120)± √(1,120)2−4⋅−16⋅(8

2 ⋅ −16

Simplify.

t=−1,120± √1,254,400+512

- 32

t= −1,120± √1,254,912

-32

Rewrite to show two solutions.

t= −1,120+ √ 1,254,912 . t= −1,120+ √ 1,254,912

-32 - 32

t≈70 seconds,t≈−0.007 seconds

Answer 2

Bullet will hit the ground after 70 seconds.

Explanation:

A bullet is fired straight up from a BB gun with initial velocity 1120 ft/s at an initial height of 8 ft. Using the value of velocity the equation becomes:

h(t)= -16t² + 1120t + 8

We need to find time when bullet hit the ground.

As we know when bullet hit the ground height would be 0

So, we set h=0 and solve for t .

0 = -16t² + 1120t + 8

Using quadratic formula:

`t= \frac{-1120 \pm \sqrt{(1120)^(2)-4(-16)(8)} }{2(-16)}\\\\ t=70.007 , -0.007`

Since negative value of the time is not possible, we conclude that the bullet will hit the ground after 70 seconds.