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.
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
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.
To find the fall time, you can use the formula for distance traveled during free fall:
S = 1/2 * g * t^2
where S is the distance traveled, g is the acceleration due to gravity (approximately 32 ft/s^2), and t is the time.
In this case, you want to find the time it takes for the ball to fall 144 ft. So you can set up the equation as follows:
144 = 1/2 * 32 * t^2
Simplifying the equation:
288 = 32 * t^2
Dividing both sides by 32:
9 = t^2
Taking the square root of both sides:
t = ±3
Since time cannot be negative in this context, the fall time is 3 seconds.
To find the fall time, we can use the formula for the height of the ball at any given time: H = -16t^2 + 64t + 80. Since the ball is falling back to the ground, the height is 0.
So we can set the equation to 0 and solve for t: -16t^2 + 64t + 80 = 0.
To solve this quadratic equation, you can use the quadratic formula: t = (-b ± √(b^2 - 4ac)) / (2a), where a, b, and c are the coefficients in the equation.
In this case, a = -16, b = 64, and c = 80. Plugging these values into the quadratic formula, we get:
t = (-64 ± √(64^2 - 4*(-16)*80)) / (2*(-16))
Simplifying further:
t = (-64 ± √(4096 + 5120)) / (-32)
t = (-64 ± √9216) / (-32)
t = (-64 ± 96) / (-32)
This gives us two possible values for t: (-64 + 96) / -32 = 32 / -32 = -1, and (-64 - 96) / -32 = -160 / -32 = 5.
Since time cannot be negative in this context, we discard -1 as a solution. Therefore, the fall time, Tf, is 5 seconds.