Calculus I Anti-derivatives Question A ball is shot from the ground straight up into the air with initial velocity of 49 ft/sec. Assuming that the air resistance can be ignored, how high does it go?

Question

A ball is shot from the ground straight up into the air with initial velocity of 49 ft/sec. Assuming that the air resistance can be ignored, how high does it go? Hint:The acceleration due to gravity is 32 ft per second squared.

Answer 1

Two integrals and one derivative.

Nice problem!

Answer 2

a(t) = acceleration at time t = -32

v(t) = velocity at time t v'(t) = a(t) So, v(t) = ∫ (-32)dt = -32t + C.

Since v(0) = 49, we have v(t) = -32t + 49.

s(t) = height at time t. s'(t) = v(t). so s(t) = ∫v(t)dt = ∫(-32t + 49)dt = -(1/16)t² + 49t + C₁

Since s(0) = 0, we have s(t) = -(1/16)t² + 49t.

At the maximum height, v(t) = 0. So, t = 49/32.

Max height = s(49/32)