Using only calculus knowledge no physics shortcuts

Question

A rock is thrown upward with a speed of 48 feet per second from the edge of a cliff 400 feet above the ground. What is the speed of the rock when it hits the ground? Use acceleration due to gravity as –32 feet per second squared and approximate your answer to 3 decimal places.

With the work I done I got 167.045 feet/sec

Answer 1

1. Velocity is the integral of acceleration
2. ∫-32dt = -32t + v0 where v0 = 48 ft/s. So velocity = 48 - 32t
3. Position is the integral of velocity
4. ∫(48-32t)dt = 48t - 16t² + x0 where x0 = 400 ft. So position = 48t - 16t² + 400
5. Position is 0 when the rock hits the ground. So set the position equation equal to 0 and solve for t.
6. - 16t² + 48t + 400 = 0
7. quadratic formula: [-48 ± √(48²-4*(-16)*400)] / (2*-16) = -3.72 seconds or 6.72 seconds
8. Plug t=6.72 seconds into the velocity equation from step 1:
9. 48 - 32(6.72) = -167.04 ft/s
10. Since the problem asks for speed, take the absolute value to get an answer of 167.04 ft/s

TL;DR: you are correct!

Answer 2

-32t for velocity is -16t^2 for position (take the anti derivative).

The position of the rock is P = -16t^2 + 48t + 400. Solve for t when it hits the ground, position = 0.

-16t^2 + 48t + 400 = 0

16t^2 -48t - 400 = t^2 - 3t - 25 = 0

t = 3 +/- sqrt (9 + 100) all divided by 2

t = 3 +/- sqrt (109) all divided by 2 . The negative one makes no sense, so it's t=(3 + sqrt(109))/2

Velocity is the derivative of position: -32t + 48. When t=0, it hits the ground.

-32((3 + sqrt(109))/2) + 48 = about -167.045

Speed is 167.045 toward the ground.