A catapult launches a cat from ground level to a target located on the ground. The cat leaves the
catapult moving at 100 km/h at an angle of 40º.
a) What is the cat’s velocity half-way through the trajectory (size and direction)?
b) What is the cat’s velocity at impact (size and direction)?
c) How long is the cat in the air for?
d) What is the maximum height?
e) What is the range of the cat?
100km/h = 100,000m/3600s = 27.78 m/s
Vo = 27.78m/s[40o],
Xo = 27.78*Cos40 = 21.28 m/s,
Yo =27.78*sin40 = 17.86 m/s,
a. At half-way point(max. ht.), Y = 0,
V = Xo + Yi = 21.28 + 0i = 21.28 m/s[0o].
b. The velocity at impact is equal to the initial velocity:
V = Vo = 27.78 m/s[40o].
c. Y = Yo + g*Tr = 0,
17.86 + (-9.8Tr = 0,
Tr = 1.82 s. = Rise time.
Tf = Tr = 1.82 s. = Fall time.
Tr + Tf = 1.82 + 1.82 = 3.64 s. = Time in air.
d. h = 0.5g * Tf^2 = 4.9 * 1.82^2 =
e. Range = Vo^2*sin(2A)/g. A = 40o, Vo = 27.78 m/s.
To solve this problem, we'll need to break down the cat's motion into horizontal and vertical components. Let's start by finding the horizontal and vertical velocities.
Given:
Initial velocity (v0) = 100 km/h
Launch angle (θ) = 40º
a) To find the cat's velocity halfway through the trajectory, we need to find the horizontal and vertical velocities at that point.
Horizontal velocity (Vx):
The horizontal velocity remains constant throughout the entire motion. It can be calculated using the initial velocity and the launch angle:
Vx = v0 * cos(θ)
= 100 km/h * cos(40º)
Vertical velocity (Vy):
The vertical velocity changes due to the acceleration acting in the downward direction (gravity). At the peak of the trajectory, the vertical velocity is zero. Therefore, halfway through the trajectory, the vertical velocity will be half of the initial vertical velocity.
Vy = v0 * sin(θ)
= 100 km/h * sin(40º)
b) The cat's velocity at impact is the final velocity just before it hits the ground. Since the target is on the ground, the vertical velocity at impact will be the negative of the initial vertical velocity.
Vy at impact = -v0 * sin(θ)
= -100 km/h * sin(40º)
c) To find how long the cat is in the air for, we need to calculate the time of flight.
Time of flight (t):
The time of flight can be found using the vertical motion equation:
0 = v0 * sin(θ) * t + (1/2) * g * t^2
In this equation, g represents the acceleration due to gravity, which is approximately 9.8 m/s^2.
Let's convert v0 to m/s and solve for t:
v0 = 100 km/h * (1000 m/1 km) * (1 h/3600 s)
= 27.78 m/s
0 = 27.78 * sin(40º) * t + (1/2) * 9.8 * t^2
Simplifying the equation, we get a quadratic equation:
4.9t^2 + 13.89t = 0
This equation can be solved using the quadratic formula or factoring:
t(4.9t + 13.89) = 0
Either t = 0 (which is the starting point) or 4.9t + 13.89 = 0
Solving the second equation for t, we get:
4.9t = -13.89
t = -13.89/4.9
t ≈ -2.83 seconds (Ignoring the negative value since time cannot be negative)
Therefore, the cat is in the air for approximately 2.83 seconds.
d) To find the maximum height reached by the cat, we can use the vertical motion equation:
h = v0^2 * sin^2(θ) / (2 * g)
Substituting the given values:
h = (27.78)^2 * sin^2(40º) / (2 * 9.8)
Simplifying, we get:
h ≈ 113.4 meters
Therefore, the maximum height reached by the cat is approximately 113.4 meters.
e) To find the range of the cat, we can use the horizontal motion equation:
Range (R) = Vx * t
Substituting the values of Vx and t:
R = (100 km/h * cos(40º)) * 2.83 seconds
Converting km/h to m/s and simplifying, we get:
R ≈ (100 * (1000/3600)) * cos(40º) * 2.83
R ≈ 173.13 meters
Therefore, the range of the cat is approximately 173.13 meters.
To answer these questions, we can break down the problem into two components: horizontal motion (x-direction) and vertical motion (y-direction).
a) To find the cat's velocity halfway through the trajectory, we need to determine the horizontal and vertical components separately.
The horizontal component of velocity remains constant throughout the entire motion. Therefore, halfway through the trajectory, the horizontal velocity component is still 100 km/h (100 km/h * cos(40°)).
The vertical component of velocity changes due to the acceleration of gravity. To find the vertical component halfway through the trajectory, we need to split the initial velocity of 100 km/h (100 km/h * sin(40°)) into its vertical and horizontal components.
Since the cat is at the highest point halfway through the trajectory, its vertical velocity component is zero at this point. Therefore, the cat's velocity halfway through the trajectory is 100 km/h (horizontal velocity component) and 0 km/h (vertical velocity component).
b) At impact, the cat's velocity will have a horizontal component and a vertical component. The horizontal component remains unchanged at 100 km/h (100 km/h * cos(40°)).
The vertical component at impact can be found by considering that the time it takes for the cat to reach the target on the ground is the same as the time it takes for the cat to reach the highest point in its trajectory and then come back down. Using kinematic equations, we can find the time of flight (T).
The initial vertical velocity component is given by 100 km/h * sin(40°). The final vertical velocity component is the negative of the initial velocity component (as the cat comes back down).
Using the formula vf = vi + at, where vf = 0, we can find the time of flight (T):
0 = (100 km/h * sin(40°)) - (9.8 m/s^2 * T)
Solving for T, we find T = (100 km/h * sin(40°)) / 9.8 m/s^2.
Substituting the value of T back into the formula, we can find the vertical component at impact.
c) The time of flight (T) is the total amount of time the cat is in the air. We have already found T in the previous step.
d) To find the maximum height, we can use the formula for vertical displacement (Δy) at the highest point of the trajectory. The vertical displacement will be the difference between the initial and final vertical positions.
The initial vertical position is zero, and the final vertical position at the highest point is given by:
Δy = (viy^2) / (2g)
where viy = initial vertical velocity component and g = acceleration due to gravity.
Plugging in the values, we get:
Δy = (100 km/h * sin(40°))^2 / (2 * 9.8 m/s^2)
e) To find the range of the cat, we can use the formula for horizontal displacement (Δx), which is the horizontal distance covered by the cat.
Since the horizontal velocity component remains constant throughout the motion, we can use the formula:
Δx = vx * T
where vx = horizontal velocity component and T = time of flight.
Plugging in the values, we get:
Δx = (100 km/h * cos(40°)) * (100 km/h * sin(40°)) / 9.8 m/s^2
By following these steps, you can find the answers to each part of the problem.