Precal.

If object is thrown upward with an intitial velocity of 15 ft per second, then its height is given by h=-15t^2 + 60t. What is the maximum height?

Two ways to find it.
1) AT the top, max height, will be one half of the time it would come back down and hit the earth at h=0
0=-15t^2+60t solve for t.
Then divide that by two.

2) At max height, the change in height/per time is zero.

dh/dt= -30t+60
set that to zero, and solve for t.

If you don't know derivatives yet, forget this method.

If object is thrown upward with an intitial velocity of 15 ft per second, then its height is given by h=-15t^2 + 60t. What is the maximum height?

The time to reach maximum height is given by Vf = Vo - gt where Vf = the final velocity, Vf = 0, Vo = the initial velocity, g = acceleration due to gravity, ~32), and t = the time.

Therefore, 0 = 15 - 32t making t = .46876 sec. sec.

The height reached is given by h = Vot - 16t^2. You give an equation of h = 60t - 15t^2 implying that Vo = 60, not 15, and g = 30, not 32.

Using your equation, h = 60(.46875) - 15(.46875)^2 = 24.829 ft. ft.

Using the initial velocity stated, h = 15(.46875) - 15(.46875)^2 = 3.735 feet.

If the real initial velocity is 15 fps, then 60 does not belong in the equation for height, rather 15 instead..

I believe this result derives from using the wrong value for g which should be 16.

Using g == 16 and the correct V0 = 15 in the equation for height, we get

0 = 15 - 16t making t = ..9375 sec.

Then, h = 15(.9375) - 16(.9375)^2 = 14.0625 - 14.0625 = 0.





This leads to h = 15(1) - 15(1) = 0
Therefore, with Vo = =15

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asked by trellis

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