I need this as fast as possible. So confused !

Question

Ball is launched vertically upward from ground level with an initial velocity of 64 feet per second then it’s height after t seconds is given by the equation h(T)=16t^2+64t 

 

how long will it take to come back down ?

after how many seconds will it be 40 fret off the ground ?

how long will it take to reach maximum height ? 

What is max height ? 

 

Answer 1

Hello Cidney,

First, I believe there should be a negative sign in front of the squared term, as gravity is working against the ball's upward motion.

 

a) To find the time at which the ball comes back to the ground, you want to set the equation equal to zero,

→ h(t) = -16t² + 64t = 0

→ -16t * (t - 4) = 0

→ t = 0 or t = 4.

 

t = 0 is the time at which the ball was initially launched,

→ we want t = 4.

 

So the ball took 4 seconds to come back down after initially being launched.

 

b) To find when it will be 40 feet off the ground, set the equation equal to 40,

→ h(t) = -16t² + 64t = 40

→ 16t² - 64t + 40 = 0

→ 2t² - 8t + 5 = 0

→ t ≈ .775, 3.225 [using the quadratic formula]

 

So after about .775 seconds on its way up, it will pass 40 feet off the ground,

and on its way back down, about 3.225 seconds after initial launch, it will pass through 40 feet again.

 

c) To find when the maximum height occurs, use the equation t = (-b) / (2a)

→ t = (8) / (2 * 2)

→ t = 2 seconds

 

So after 2 seconds, the maximum height will be reached.

 

d) To find the max height, plug t = 2 into h(t)

→ h(2) = -16(2)² + 64(2)

→ h(2) = -64 + 128

→ h(2) = 64 feet

 

So the max height is 64 feet.

 

I hope the above helps, and thank you for the question.

Michael E.