What time does the ball hit the ground? I think we must use antiderivative.

Question

A ball is thrown upward with a speed of 20 ft/s from the edge of a cliff that is 1400 ft above the ground. At what time t does the ball hit the ground?

(Recall that acceleration due to gravity is 32 ft/s2. Round your answer to two decimal places or enter an exact answer.)

Answer 1

Recall the relationship between distance, s(t), speed, v(t), and acceleration, a(t).

s'(t) = v(t) and s''(t) = v'(t) = a(t),

which implies that v(t) is the antiderivative of a(t), and s(t) is the antiderivative of v(t).

From the problem, we know that at time t=0, s(0) = 1400 ft and v(0) = 20 ft/s. We are given the acceleration, a constant -32ft/s². Since v(t) is an antiderivative of a(t), we have v(t) = ∫ a(t)dt = -32t + C, where C is some constant. However, we have an initial condition for v, v(0) = 20 ft/s. Hence, 20 = v(0) = -32(0) + C = C. In particular, notice that our constant of integration is just v(0)!. Hence, v(t) = -32t+20.

Use a similar process to find s(t), remembering s(t) is an antiderivative of v(t)! Once you've found s(t), solve the equation s(t) = 0 to find when the ball hits the ground. Hint: Quadratic formula and then use some reasoning to decide what solution fits the context of the problem.