A rock thrown straight up with a velocity of
29 m/s from the edge of a building just misses
the building as it comes down. The rock is
moving at 44 m/s when it strikes the ground.
How tall was the building? The acceleration of gravity is 9.8 m/s
Answer in units of m
To find the height of the building, we need to consider the motion of the rock both when it is thrown up and when it comes back down.
First, let's find the time it takes for the rock to reach its maximum height. We can use the formula:
v = u + at
where:
v = final velocity (0 m/s when the rock reaches its maximum height)
u = initial velocity (29 m/s)
a = acceleration (-9.8 m/s², considering the negative direction is up)
t = time
Rearranging the formula to solve for 't':
0 = 29 - 9.8t
Simplifying further:
9.8t = 29
t = 29 / 9.8
t ≈ 2.96 s
Therefore, it takes approximately 2.96 seconds for the rock to reach its maximum height.
Next, we can use the formula to find the maximum height (h) reached by the rock:
h = ut + (1/2)at²
where:
h = maximum height
u = initial velocity (29 m/s)
t = time (2.96 s)
a = acceleration (-9.8 m/s²)
Plugging in the values:
h = 29(2.96) + (1/2)(-9.8)(2.96)²
h = 85.84 - 43.856
h ≈ 41.984 meters
Therefore, the maximum height reached by the rock above the edge of the building is approximately 41.984 meters.
To find the height of the building, we need to consider the height from the maximum height to the ground. The rock will fall from this maximum height with a final velocity of 44 m/s.
Using the formula:
v² = u² + 2ah
where:
v = final velocity (44 m/s)
u = initial velocity (0 m/s when the rock reaches its maximum height)
a = acceleration (-9.8 m/s²)
h = height
Since the initial velocity (u) is 0 m/s, the equation simplifies to:
v² = 2ah
Plugging in the values:
44² = 2(-9.8)h
h = (44²) / (2(-9.8))
h ≈ 96.69 meters
Therefore, the height of the building is approximately 96.69 meters.