# Calculus Velocity and Acceleration Problem.

Okay so this is the problem: "On a certain planet, a ball is thrown upward with a speed of 33 feet per second from the edge of a cliff 44 feet above the ground. When does it hit the ground? (use a(t) = −22).

I already know that the answer is 4 seconds but I don't understand how to get there. Can someone please guide me and explain to me the process of solving this problem? Thank you!

## 1 Answer

Hint: Start with the fact that acceleration is the derivative of velocity, which is the derivative of position. We know that $a(t) = -22$, so integrate it to find the velocity, using the information about the initial velocity in order to find the constant of integration. Then once you've found the velocity $v(t)$, you can integrate that to find the position, and again you'll have a new constant of integration that you can find the value of by using the information about the original position. Once you've found the position $x(t)$, you can solve for the time $t$ where the position of the ball is at the ground, i.e. $x = 0$.

• Okay, so I integrated the acceleration function to get v(t) to be -22t. How exactly do I go about finding the constant? – Dylan Apr 23 '14 at 23:06
• Would it v(t) be -22t + 33? – Dylan Apr 23 '14 at 23:07
• @Dylan Yes it would, because when you integrate you get $v(t) = -22t + C$, and then you use the fact that $v(0) = 33$ to solve for $C$, and you get $C=33$. – Keshav Srinivasan Apr 23 '14 at 23:38
• So I'm assuming s(t) = -11t^2 + 33t + 44? – Dylan Apr 23 '14 at 23:41
• Okay, so I used the position function I stated above, factored it out, and found 4 to be a zero as well as -1. Why is 4 the answer but not -1? – Dylan Apr 23 '14 at 23:43