Answer:
Bullet will hit the ground after 70 seconds.
Step-by-step 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. Â
As we know when bullet hit the ground height would be 0 Â
So, we set h=0 and solve for t .
0 = -16t² + 1120t + 8
Using quadratic formula:
[tex]t= \frac{-1120 \pm \sqrt{(1120)^{2}-4(-16)(8)} }{2(-16)}\\\\ t=70.007 , -0.007[/tex]
Since negative value of the time is not possible, we conclude that the bullet will hit the ground after 70 seconds.
Answer:
t≈70 seconds
Step-by-step 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
