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
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
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.