Find the initial velocity of the second ball

[gigi9]

Calc help please!

Can someone please help me how to do this problem using calculus? Thanks a lot.
A ball is dropped out of a window 19.6m above the ground. At the same time another ball is thrown straight down from a window 79.6m above the ground. If both balls reach the ground at the same moment, find the initial velocity of the second ball.
***position = s(t)= -16t^2 + ct +d

 

[PrudensOptimus]

v(t) = s'(t) = -32t + c

Since they reach the ground, v(final) = 0, you find v(t) = 0.

-32t + c = 0.

t = c/32 s.

The time at which the ball reaches 0 m/s is at c/32 s. It takes the ball c/32 s to reach it's highest point before start falling back to earth. Since the time which the ball reaches the ground is equal to the time it takes to reach the highest point, the answer should be c/32 s.

the velocity before it hits the ground is v(c/32) = -ct + c = c(c - t).

 

[HallsofIvy]

v(t) = s'(t) = -32t + c

Since they reach the ground, v(final) = 0, you find v(t) = 0.

-32t + c = 0.

t = c/32 s.

What? The velocity of a dropped object or one thrown downward is NOT 0 when it hits the ground! Of course, you change that later. What you are doing is finding the highest point (for an object thrown UPWARD) to find the initial velocity of the object. Assuming that "hitting the ground" means coming back to its initial height, that would be the same as the final speed. Normally you do better than this, Prudens Optimus. Did you misread the problem?



NONE of those are true here! One object is dropped (initial velocity 0) and the other is thrown downward (initial velocity unknown- call it c as gigi9 did.
The object dropped has height h(t) at time t given by h(t)= -9.8t²+ 19.6 (height is measured in meters from the ground, t in seconds).
The object thrown downward has height h(t)= -9.8t²+ ct+ 79.6.

Solve -9.8t²+ 19.6= 0 (h= 0 when the object hits the ground) for t, then put that value of t into -9.8t²+ ct+ 79.6= 0 and solve for c (it will be negative, of course).