In a scene in an action movie a person is being chased across a roof top. To escape, the person runs off the edge of the building and falls to the roof of a nearby shorter building. The tall building is 50m high, the short building is 40m high and there is a 5m gap in-between the buildings.
a. Draw a vector diagram to describe this situation.
b. Ignoring air resistance, what is the minimum velocity the person must be running at to safely land on the shorter building?
c. What is the magnitude of the velocity of the person when they land on the smaller building?
HELP!!
a. To draw a vector diagram for this situation, you can represent the person's initial velocity, the gravitational force acting on them, and their final velocity when they land on the shorter building.
First, draw a vertical line to represent the tall building. Label it as the initial position of the person. Then, draw a horizontal line to represent the smaller building, starting from the end of the vertical line. Label it as the final position of the person.
Next, draw a downward arrow representing the force of gravity acting on the person. Label it as the weight or gravitational force. The length of this arrow can be arbitrary, as we do not need to determine its precise magnitude in this case.
Finally, draw an arrow starting from the initial position of the person and ending at the final position. This arrow represents the person's velocity vector. Since the only force acting on the person is gravity, the vector's direction should be straight down. The magnitude of this vector will be determined in parts b and c.
b. To determine the minimum velocity the person must be running at, we need to consider the time it takes for them to fall from the tall building and cross the gap.
First, calculate the time it takes for the person to reach the ground from the tall building using the equation:
h = (1/2)gt^2
Where:
h = height of the tall building = 50m
g = acceleration due to gravity = 9.8 m/s^2
t = time taken to fall
Rearranging the equation, we have:
t = √(2h/g)
t = √(2 * 50 / 9.8)
t ≈ 3.19s
Now, calculate the horizontal distance that needs to be covered during this time:
d = vt
Where:
d = horizontal distance = 5m
v = horizontal velocity (required velocity)
Rearranging the equation, we have:
v = d / t
v = 5 / 3.19
v ≈ 1.57 m/s
Therefore, the minimum velocity the person must be running at to safely land on the shorter building is approximately 1.57 m/s.
c. To find the magnitude of the velocity of the person when they land on the shorter building, we can use the principle of conservation of energy. At the point of landing, the potential energy on the tall building is converted into kinetic energy on the shorter building.
The potential energy (PE) at the top of the tall building is given by:
PE = mgh
Where:
m = mass of the person (which we will assume to be 1 kg for simplicity)
g = acceleration due to gravity
h = height of the tall building = 50m
The kinetic energy (KE) on the shorter building is given by:
KE = (1/2)mv^2
Where:
m = mass of the person = 1 kg
v = velocity of the person when they land on the shorter building
Since energy is conserved, we can equate the potential energy to the kinetic energy:
mgh = (1/2)mv^2
Simplifying the equation:
gh = (1/2)v^2
v = √(2gh)
Substituting the known values:
v = √(2 * 9.8 * 40)
v ≈ 19.8 m/s
Therefore, the magnitude of the velocity of the person when they land on the smaller building is approximately 19.8 m/s.