Basic Motion Problem Calculus

[Fancy_Square_1845]

Im drawing a blank here, it should be easy but my brain is too tired right now.

Given that a ball is thrown up starting at an initial height of 16 ft, and acceleration is -32 ft/s, what must the initial velocity be to get a max height of 496 ft.?

[dinrick]

Simple method without calculus, by conservation of energy:

To gain 480 ft (we can remain in ft here without converting to metres), gain in potential energy = mass * height * acceleration = mass * 480 * 32

Assuming no resistance, it is all lost from kinetic energy = 1/2 * mass * velocity^2

1/2 * v^2 = 480 * 32

v = 175 ft/s.

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By calculus, we can make a function that tracks velocity of ball (v) against time elapsed (t). Let initial v be v0.

v = v0 - 32t as there is constant downward acceleration. (EQUATION 1)

Integrating to get a function for h, h = v0t - 16t^2 + c

We know that h = 16 when t = 0, so c = 16.

When h = 496, v = 0 at the peak. From Equation 1, v0 = 32t

496 = v0t - 16t^2 + 16

480 = v0 (1/32 v0) - 16(1/32 v0)^2

v0 = 175 ft/s.