An object is thrown downward with an initial
speed of 15 m/s from a height of 85 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.
just calculate their heights at time t. You want
85-15t-4.9t^2 = 29t-4.9t^2
The graphs shown here might help visualize things:
http://www.wolframalpha.com/input/?i=85-15t-4.9t^2+%3D+29t-4.9t^2
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 point.
We can start by finding the time it takes for the first object (thrown downward) to reach the ground. We can use the equation:
๐ = ๐ฃโ๐ก + 1/2๐๐กยฒ
where ๐ is the initial height of 85m, ๐ฃโ is the initial downward velocity of 15 m/s, ๐ is the acceleration due to gravity (-9.8 m/s^2), and ๐ก is the time taken.
Plugging in the values:
85 = 15๐ก + 1/2(-9.8)๐กยฒ
Simplifying, we get:
4.9๐กยฒ + 15๐ก - 85 = 0
This is a quadratic equation that we can solve using the quadratic formula:
๐ก = (-๐ ยฑ โ(๐ยฒ - 4๐๐)) / 2๐
where ๐ = 4.9, ๐ = 15, and ๐ = -85.
Calculating ๐ก using the quadratic formula, we find two values: ๐กโ and ๐กโ. Since we are only interested in the time it takes for the object to reach the ground, we can ignore the negative value ๐กโ.
Now that we have ๐กโ, we can find the height at which the first object and the second object pass each other.
The position of the first object at time ๐กโ can be calculated using the equation:
โโ = ๐ + ๐ฃโ๐ก + 1/2๐๐กยฒ
Substituting the values:
โโ = 85 + 15๐กโ + 1/2(-9.8)๐กโยฒ
Finally, we substitute the value of ๐กโ that we found earlier into this equation to get the height at which the two objects pass each other.