A rock thrown straight up with a velocity of 30 m/s from the edge of a building just misses the building as it comes down. The rock is
moving at 54 m/s when it strikes the ground. How tall was the building? The acceleration of gravity is 9.8 m/s^2
Answer in units of m.
To find the height of the building, we can break down the problem into two parts: when the rock is thrown up and when it comes back down.
1. When the rock is thrown up:
Let's consider the upward motion first. We know the initial velocity (u) is 30 m/s, and the final velocity (v) is 0 m/s because the rock reaches its highest point where it momentarily stops before falling.
Using the equation of motion:
v^2 = u^2 + 2as
we have:
0^2 = 30^2 + 2(-9.8)s
Simplifying the equation:
0 = 900 - 19.6s
Rearranging the equation:
19.6s = 900
Solving for 's':
s = 900 / 19.6 ≈ 45.92 m
The height when the rock is thrown up is approximately 45.92 m.
2. When the rock falls down:
Let's consider the downward motion now. We know the initial velocity (u) is 0 m/s (the rock starts falling from rest), and the final velocity (v) is -54 m/s (negative because it falls downward).
Again using the equation of motion:
v^2 = u^2 + 2as
we have:
(-54)^2 = 0^2 + 2(-9.8)s
Simplifying the equation:
2916 = -19.6s
Dividing both sides by -19.6:
s = 2916 / (-19.6) ≈ -148.57 m
The height when the rock falls down is approximately -148.57 m, but we know that the height cannot be negative. Therefore, we take the absolute value of this height:
|s| = |-148.57| = 148.57 m
So, the height when the rock comes down is approximately 148.57 m.
Now, to find the total height of the building, we combine the heights when the rock is thrown up and when it comes down:
Total height = height when thrown up + height when coming down
= 45.92 m + 148.57 m
≈ 194.49 m
Therefore, the height of the building is approximately 194.49 m.