Calculus BC Word Problem NEED HELP ASAP PLS!

Question

An elementary student kicks a ball straight into the air with a velocity of 16 feet/sec. If acceleration due to gravity is -32 ft/sec², how many seconds after it leaves his foot will it take the ball to reach its highest point? Assume the position at time t = 0 is 0 feet. 

Answer 1

a = dV/dt so dV = adt. Integrating both sides we get:

∫dV = ∫adt or

V = at + C₁ but since we are given the initial velocity (t = 0) is 16 ft/sec, then C₁ = 16 giving us:

V = -32t + 16

dy/dt = V. Subbing that into the above, we have:

dy/dt = -32t + 16 or

dy = (-32t + 16)dt and integrating both sides we get:

∫dy = ∫(-32t + 16)dt or

y = -32/2t² + 16t + C₂

y = -16t² + 16t + C₂ and since we are given the initial position (t = 0) is 0, then C₂ = 0

So, y = -16t² + 16t

This parabola has a max height at the vertex. The vertex occurs at t = -b/(2a) = -16/(2(-16)) = 1/2

So the time for the max height is 1/2 second.

BTW, at t = 1/2, the height is y = -16(1/2)² + 16(1/2) = -4 + 8 = 4 feet