Calc Antiderivative Help

Question

A hot air balloon is rising vertically with a velocity of 2.0 feet per second. A very small ball is released from the hot air balloon at the instant when it is 1120 feet above the ground. Use a(t)=−32 ft/sec^2 as the acceleration due to gravity. 

1. How many seconds after its release will the ball strike the ground? Round the answer to the nearest two decimal places. 

2. At what velocity will it hit the ground? 

Answer 1

This problem can be solved using the CAKE equations (Constant Acceleration Kinematic Equations)

In particular h(t) = h₀ + v₀ t + (1/2) a t²

Here h₀ = 1120 , v₀ = 2 and a = - 32

The time before the ball strikes the ground is the time-of-flight , t_tof

h(t_tof) = 0 ( ground is at height zero) This gives a quadratic equation to solve :

0 = 1120 + 2 t_tof - 32 (t_tof)²

I used the quadratic equation to solve this getting t_tof = 5.947 s

For part 2 we use the equation v = v₀ + a t

v(t_tof) = 2 - 32 x 5.947 = -188.3 ft/s ( negative because going down)