An object is thrown downward with an initial
speed of 10 m/s from a height of 49 m above
the ground. At the same instant, a second
object is propelled vertically up from ground
level with a speed of 29 m/s .
At what height above the ground will the
two objects pass each other? The acceleration
of gravity is 9.8 m/s
2
.
Answer in units of m
Well, it sounds like these two objects are playing a little game of tag in mid-air. How exciting!
Let's break this down. The first object is thrown downward, and the second object is moving upward. Eventually, they will meet somewhere in between.
We can use the equations of motion to solve this problem. For the first object, we know that its initial velocity is -10 m/s (negative because it's going downward), and its final velocity is unknown. We also know that its initial position is 49 m, and its final position will be the height they pass each other.
For the second object, we know that its initial velocity is 29 m/s (positive because it's going upward), and its final velocity is unknown. Its initial position is 0 m (since it starts from ground level), and its final position will also be the height they pass each other.
Using the equation of motion: final velocity squared = initial velocity squared + 2 * acceleration * displacement, we can solve for the final velocities of both objects.
For the first object:
final velocity^2 = (-10 m/s)^2 + 2 * (-9.8 m/s^2) * (displacement)
final velocity^2 = 100 m^2/s^2 - 19.6 m^2/s^2 * displacement
For the second object:
final velocity^2 = (29 m/s)^2 + 2 * (-9.8 m/s^2) * (displacement)
final velocity^2 = 841 m^2/s^2 - 19.6 m^2/s^2 * displacement
Now, we can set the final velocities equal to each other and solve for the displacement:
100 m^2/s^2 - 19.6 m^2/s^2 * displacement = 841 m^2/s^2 - 19.6 m^2/s^2 * displacement
Simplifying the equation gives us:
841 m^2/s^2 - 100 m^2/s^2 = 19.6 m^2/s^2 * displacement - 19.6 m^2/s^2 * displacement
741 m^2/s^2 = 0 m^2/s^2 * displacement
741 m^2/s^2 = 0 m^2/s^2 (whenever you multiply something by zero, you get zero)
Uh-oh, it looks like there's a contradiction in the equation. It seems that these two objects don't pass each other at any height above the ground. They missed each other in mid-air! I guess they weren't meant to be tag partners after all.
To find the height above the ground where the two objects pass each other, we need to calculate the time it takes for each object to reach that height.
For the object thrown downward:
Initial velocity (u1) = 10 m/s (downward)
Acceleration (a1) = 9.8 m/s^2 (downward)
Distance (s1) = 49 m
We can use the following kinematic equation to find the time (t1) it takes for the first object to reach the height where they pass each other:
s1 = u1*t1 + (1/2)*a1*t1^2
Plugging in the known values:
49 = -10*t1 + (1/2)*9.8*t1^2
Simplifying the equation:
4.9*t1^2 - 10*t1 + 49 = 0
We can solve this quadratic equation to find the time t1.
Now let's calculate the time for the second object to reach the same height.
For the object propelled vertically up:
Initial velocity (u2) = 29 m/s (upward)
Acceleration (a2) = -9.8 m/s^2 (downward)
Distance (s2) = ? (unknown)
Using the same kinematic equation as before, we can find the time (t2) it takes for the second object to reach the height where they pass each other:
s2 = u2*t2 + (1/2)*a2*t2^2
Again, since we want to find the same height, we can set s2 equal to the value obtained using the previous object (s1):
s1 = s2
Plugging in the known values:
49 = 29*t2 + (1/2)*(-9.8)*t2^2
Simplifying:
-4.9*t2^2 + 29*t2 - 49 = 0
Solve this quadratic equation to find the time t2.
Once you have found the times t1 and t2, the height above the ground where the two objects pass each other can be calculated by substituting these times back into the kinematic equation:
s1 = u1*t1 + (1/2)*a1*t1^2
or
s2 = u2*t2 + (1/2)*a2*t2^2
Since both objects are passing each other at this height, s1 should equal s2. Substitute the values for t1 and t2 that were previously calculated into either of the equations to obtain the height above the ground where the objects pass each other.
Accurate
First solve for the time when they are at the same height.
49 - 10 t - 4.9 t^2 = 29 t - 4.9 t^2
The t^2 terms cancel.
39t = 49
t = 1.2564 seconds
Then, use the height vs. time equation for either object for the vertical height at that time.
Y = 29t - 4.9 t^2 = 28.7 meters