Frank stands at the base of a building. He throws a 0.12kg rock straight up at a speed of 28m/s so that it rises to a maximum height of 12m above the top of the building and then falls straight down on the roof of the building.
a) What is the height of the building?
b) What is the velocity of the rock just before it lands on the roof?
a) Well, if the maximum height of the rock above the top of the building is 12m, then the height of the building is...wait for it...12m! You can't underestimate the power of deduction.
b) Now, the velocity of the rock just before it lands on the roof can be calculated using some clever physics. Since the rock is falling straight down, we know that it will be moving downwards with the same speed it had when it was thrown up. So, the velocity just before it lands on the roof is...minus 28m/s! It's like the rock was in a real hurry to get back down.
To determine the height of the building and the velocity of the rock just before it lands on the roof, we can use the equations of motion.
a) To calculate the height of the building, we need to find the total distance the rock traveled. This is the sum of the distance it rose above the top of the building and the distance it fell from that point to the roof.
Let's break down the problem into two parts:
1. The ascent: The rock is thrown straight up with an initial velocity of 28 m/s and reaches a maximum height of 12 m above the top of the building. Here, the acceleration due to gravity acts as a decelerating force to slow down the rock.
To find the time taken to reach the maximum height, we can use the equation:
v_f = v_i + at
Since the rock reaches zero velocity at its maximum height, we can set v_f = 0 and solve for t:
0 = 28 m/s - 9.8 m/s^2 * t
Solving this equation for t gives us t = 2.857 seconds.
The height reached during the ascent can be calculated using the equation:
h = v_i*t + (1/2)*a*t^2
Substituting the known values, we have:
h = 28 m/s * 2.857 s + (1/2) * (-9.8 m/s^2) * (2.857 s)^2
h = 80 m
2. The descent: After reaching its maximum height, the rock falls back down to the roof. The distance it travels during this phase is equal to the height of the building.
Therefore, the height of the building is 80 m.
b) To find the velocity of the rock just before it lands on the roof, we can use the equation:
v_f = v_i + at
Since the rock starts from rest at the top of its trajectory, the initial velocity (v_i) is 0. We know the acceleration (a) due to gravity is -9.8 m/s^2. We need to find the final velocity (v_f) when the rock hits the roof.
Rearranging the equation, we have:
v_f = at
Substituting the known values:
v_f = -9.8 m/s^2 * 2.857 s
v_f ≈ -28 m/s
Therefore, the velocity of the rock just before it lands on the roof is approximately -28 m/s (negative sign indicates it is moving downward).