Please help if you can.
If a ball is shot projected straight upward at a rate of 64 feet per second from a rooftop, 80 feet above ground level, the height of feet above ground is modeled by the equation,
h(t)=-16t^2+64t+80
Find h(0), h(1),h(2)
Use the zero-factor property to determine how many seconds it will take for the ball to reach the ground? (What value of t makes h(t)=0)
I will do one of them, you do the rest following my method, ok?
f(1) = -16(1)^2 + 64(1) + 80
= -16 + 64 + 80
= 128
when h = 0
-16t^2+64t+80 = 0
divide by -16
t^2 - 4t - 5 = 0
(t-5)(t+1) = 0
so t=5 or t=-1
clearly t > 0 ,so after 5 seconds it will reach the ground.
Thank you so much. You really helped me understand. I can definitely finish the rest.
To find the height of the ball above the ground at different points in time, we can substitute different values of t into the equation h(t) = -16t^2 + 64t + 80.
To find h(0), substitute t = 0 into the equation:
h(0) = -16(0)^2 + 64(0) + 80
h(0) = 0 + 0 + 80
h(0) = 80 feet
To find h(1), substitute t = 1 into the equation:
h(1) = -16(1)^2 + 64(1) + 80
h(1) = -16 + 64 + 80
h(1) = 128 feet
To find h(2), substitute t = 2 into the equation:
h(2) = -16(2)^2 + 64(2) + 80
h(2) = -64 + 128 + 80
h(2) = 144 feet
Now let's determine how many seconds it will take for the ball to reach the ground. This is done by finding the value of t that makes h(t) = 0.
Set h(t) = 0 and solve for t:
-16t^2 + 64t + 80 = 0
To simplify the equation, divide the entire equation by 16:
-t^2 + 4t + 5 = 0
To solve the quadratic equation, we can factor it or use the quadratic formula. In this case, the equation can be factored as:
-(t - 1)(t - 5) = 0
So, either (t - 1) = 0 or (t - 5) = 0.
This means t = 1 or t = 5.
Therefore, it will take the ball 1 second or 5 seconds to reach the ground.