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?
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
To find the time it takes for the ball to fall back to the ground (fall time), we need to set up the equation using the formula for the height of the ball at time T:
H = -16t^2 + 64t + 80
We know that the height of the ball at the maximum height is 144 ft, which occurs at time T = 2 s. So, substituting T = 2 into the equation, we have:
144 = -16(2)^2 + 64(2) + 80
Simplifying this equation:
144 = -16(4) + 128 + 80
144 = -64 + 128 + 80
144 = 144
Now, the ball is back on the ground when its height is 0, which means we need to set H = 0 in the equation and solve for the time Tf:
0 = -16Tf^2 + 64Tf + 80
Dividing the equation by 16 to simplify it:
0 = -Tf^2 + 4Tf + 5
To solve this equation, we can factor it or use the quadratic formula. However, in this case, factoring seems to be a bit difficult. So, let's use the quadratic formula:
Tf = (-b ± √(b^2 - 4ac)) / 2a
In our equation, a = -1, b = 4, and c = 5. Substituting these values, we have:
Tf = (-(4) ± √((4)^2 - 4(-1)(5))) / 2(-1)
Simplifying the equation:
Tf = (-4 ± √(16 + 20)) / -2
Tf = (-4 ± √36) / -2
Now, simplify further:
Tf = (-4 ± 6) / -2
We have two possible solutions:
1. Tf = (-4 + 6) / -2 = 2 / -2 = -1 second (which is not possible because time cannot be negative)
2. Tf = (-4 - 6) / -2 = -10 / -2 = 5 seconds
Therefore, the fall time (Tf) is 5 seconds.