Question

Hoping Henry can help on previous prob.
A ball is thrown upwards from a roof top, 80ft 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, which is given by H= -16t2 + 64t +80.



V = Vo + g*Tr = 0.
64 + 32Tr = 0,
Tr = 2 s. = Rise time or time to reach max height, H.

H = -16*2^2 + 64*2 + 80 = 144 Ft.

0.5g*Tf^2 = 144.
16Tf^2 = 144,
Tf = 3 s. = fall time.

T = Tr + T = 2 + 3 = 5 s. = Time in flight.
Please elaborate,
How did you come up with 3s for the fall time?

Quadratic Equation - Henry - Steve today at 4:40am
max height reached at t=2
h=0 at t=5

so, 2 seconds rising, 3 seconds falling

he got the 3 by seeing how long it takes to fall 144 ft. s = 1/2 at^2


Posted my 2 previous math problem. What formula would I use and how would I apply it to get the 3 seconds? 144 ft. s = 1/2 at^2 , I must be doing something wrong cause it didn't seem to work.

Answer 1

Hf=-16t2 + 64t +80. If you have the rise time, from the velocity at top is zero, as above, you get rise time of 2 seconds.

Now, if you set Hf=0, solve the q

Answer 2

continured...solve the quadratic for t when Hf is zero, you get t=5 seconds. So, if it took five seconds for entire flight, but two seconds to go up, then it is three seconds to fall.