# Can someone help me with this problem?

1. May 4, 2004

### Buffalob07

A rock is thrown vertically with the celocity of 24 meters per second. It reaches a height of 2+24t-4.9t^2 after t seconds. How many seconds after the rock is thrown will it reach maximum height? What is the maximum height in meters? How many seconds after the rock is thrown will it hit the ground? Round answers to the nearest hundredth. Thanks.

2. May 4, 2004

position = s(t) = 2 + 24t - 4.9t^2

velocity = v(t) = s'(t) = 24 - 9.8t

The maximum height will be reached when velocity equals 0 (because once the rock's velocity reaches 0, it will be come negative and start falling downward.

24 - 9.8t = 0
9.8t = 24
t = 2.45

So, to answer the first question, maximum height is reached at t = 2.45 seconds.

Now, to find the height reached, take the original position function and subtitute in the time...

s(2.45) = 2 + 24(2.45) - 4.9(2.45)^2 = 31.39.

The maximum height reached is 31.39 meters.

Someone else's input with the last question should be used. I tried doing it and I got an undefined answer, so I'm guessing I'm doing it wrong.

3. May 6, 2004

### Gokul43201

Staff Emeritus
Continuing where IG left off... t = 2.45 s for the climb, and another 2.45s to come back down to the original position, at a height of 2 m. That's 4.9s so far. Now there's an additional 2m to go, and the velocity is now 24 m/s downwards.

from s = ut + 4.9t^2, with s = 2 and u = 24, you get,
4.9t^2 + 24t -2 = 0
Solution is t = (-24 + sqrt(576+39.2))/9.8 = 0.08 s

Total time = 4.98s