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?
To find the height of the building and the velocity of the rock just before it lands on the roof, we can use the principles of projectile motion.
a) To find the height of the building:
Since the rock rises to a maximum height of 12m above the top of the building, we can calculate the total height covered by the rock. The total height is the sum of the height from the base of the building to the top and the additional 12m it rises above the top.
Let's start by finding the time it takes for the rock to reach its maximum height. We can use the equation:
v = u + at
Where:
v = final velocity (0 m/s at maximum height)
u = initial velocity (28 m/s)
a = acceleration (due to gravity, approximately -9.8 m/s^2)
t = time
Rearranging the equation, we have:
t = (v - u) / a
t = (0 - 28) / -9.8
t = 2.857 seconds (rounded to three decimal places)
Now, we can find the height from the base of the building to the top using the equation:
s = ut + (1/2)at^2
Where:
s = distance (height)
u = initial velocity (28 m/s)
t = time (2.857 seconds)
a = acceleration (due to gravity, approximately -9.8 m/s^2)
Plugging in the values, we get:
s = 28(2.857) + (1/2)(-9.8)(2.857)^2
s = 80 meters (rounded to two decimal places)
Therefore, the height of the building is 80 meters.
b) To find the velocity of the rock just before it lands on the roof:
The velocity of the rock just before it lands on the roof can be found using the equation:
v = u + at
Where:
v = final velocity
u = initial velocity (0 m/s when the rock is at its maximum height)
a = acceleration (due to gravity, approximately -9.8 m/s^2)
t = time it takes for the rock to fall from its maximum height to the roof
The time it takes for the rock to fall from its maximum height to the roof can be found using:
s = ut + (1/2)at^2
Where:
s = distance (12 meters, the additional height the rock falls)
u = initial velocity (0 m/s)
t = time
a = acceleration (due to gravity, approximately -9.8 m/s^2)
Plugging in the values, we have:
12 = 0(π‘) + (1/2)(-9.8)(π‘)^2
Rearranging the equation gives us a quadratic equation:
4.9π‘^2 = 12
Solving for t, we find:
π‘^2 = 12 / 4.9
π‘^2 β 2.448
π‘ β 1.564 seconds (rounded to three decimal places)
Now, we can find the final velocity using the equation:
v = u + at
Plugging in the values, we get:
v = 0 + (-9.8)(1.564)
v β -15.247 m/s (rounded to three decimal places)
Therefore, the velocity of the rock just before it lands on the roof is approximately -15.247 m/s. The negative sign indicates that the velocity is in the opposite direction of the initial throw, i.e., downward.
At the very top of the flight, velocity is zero.
Vf=vi-4.9t
0=vi-4.9t
t= vi/4.2 Now put that time in
hmax= vi*t-4.9t^2
heightbuilding= hmax-12
Then, find vf when h=heightbuilding