- #1

- 23

- 0

## Homework Statement

An object is thrown downward with an initial

speed of 4 m/s from a height of 99 m above the

ground. At the same instant, a second object

is propelled vertically from ground level with

a speed of 25 m/s.

The acceleration of gravity is 9.8 m/s^2.

At what height above the ground will the

two objects pass each other? Answer in units

of m.

## Homework Equations

None - I believe you make your own equation.

Look below to see what I mean.

## The Attempt at a Solution

Well I tried to find the position equation of both

objects, then set them equal to each other to

find the time at which their positions are equal.

For the top object, since at time 0 his position

is 99 m, and his initial velocity is 4 m/s, I found

its position equation to be

**x(t) = -4t + 99**. The

-4t is negative because its traveling down.

For the bottom object, since at the time 0 his

position is also 0, it has no constant. His initial

velocity is 25 m/s, so I got his position equation

to be

**x(t) = 25t**. The 25 t is positive because he

is propelled upwards.

So I set the two equations equal, and solved for t:

-4t + 99 = 25t

99 = 29t

**t = 3.4138**

I plugged this time into one of my position

equations to find the position when they

meet:

x(t) = 25t

x(3.4138) = 25(3.4138)

**x = 85.3448 m**

This answer seemed to make sense, because

since the bottom object is propelled significantly

faster than the top object, they should meet

somewhere close to the 99 m mark.

But low and behold, 85.3448 m is incorrect.

Help!!!