Kinematics - Projectile Motion

In summary: Therefore, in summary, we can determine the initial velocity of the boulder by setting up and solving two equations using the given information and the general expression for a projectile. This will help us determine if the catapult can launch the stone within the safety guidelines and how high it will hit the wall if it does not.
  • #1
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Homework Statement



You and your friend have been hired to see if the catapult a movie company owns will be usable for a scene about a battle set in the middle ages. In the scene a catapult launches a 750 kg boulder that is supposed to hit the ground 350 m away. The director is worried the boulder will hit a 7.0 m high stone wall the movie company built 325 m from the catapult. You have measured the angle the boulder leaves the catapult relative to the ground to be 30 degrees. Because there is an airport near the movie filming location, there is a safety rule that the boulder cannot fly over 200 m high. Assume that the catapult is 2 m off the ground and the movie set is on flat ground. Can the catapult launch the stone and not violate the safety rule so that the stone clears the wall? If not, how high on the wall will it hit and if so, by how much will the boulder miss the top of the wall?


Homework Equations



d=vit+1/2at2
v22 = v12 +2ad
vix=d/t
x=(vo2sin2theta)/g
Fnet=ma

The Attempt at a Solution



My main issue with this question is the fact that the catapult is launched from 2 m above the ground but it lands on the ground. I tried using the 2 m point as the "ground" or reference line, but that didn't really help because it messes with the range (the projectile travels 350 m when it is launched from 2 m above the ground, not when it is launched from the ground. If it is launched from the ground, the range wouldn't be 350 m). This makes it difficult to find the maximum height (which must be less than 200 m to meet the safety guideline). I tried using the equation horizontal range=[(vo)2(sin2theta)]/g, but that only works if the object returns to the level it was launched at, and in this case it doesn't, so I don't know how else to find the inital velocity only being given the mass of the boulder and the angle it is launched at. I tried splitting the diagram into two halves and seeing whether I could examine the motion of the boulder during the second half (with a range of 175 m, starting from the highest point where vy=0) but this didn't work because I did not know the initial velocity. Thanks in advance for your help! :)
 
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  • #2
The general expression for a projectile is given by
y = tan(theta)*x - g*x^2/(2Vo^2*cos^2theta).
In the first case take y = -2m because it is along the negative y axis, and in the second case take y as ( 7m - 2m ) = 5m. Solve the two equations to find Vo
 
  • #3

I would first suggest that we clarify the initial conditions of the experiment. It is important to know if the 2 m height of the catapult is included in the 350 m range or if it is an additional distance. Additionally, we need to determine if the 7.0 m high stone wall is considered part of the 350 m range or if it is outside of it. This will greatly impact our calculations and results.

Assuming that the 2 m height is included in the 350 m range and the wall is outside of it, we can proceed with the following calculations:

Using the equation d=vit+1/2at^2, we can calculate the initial velocity (vi) of the boulder. Plugging in the given values of d=350 m, a=g=9.8 m/s^2, and t=3 seconds (since the boulder will take 3 seconds to reach the ground at a 30 degree angle), we get vi= 69.6 m/s.

Next, using the equation v^2=v0^2 + 2ad, we can calculate the maximum height (h) reached by the boulder. Plugging in the values of v=0 (since the boulder reaches its maximum height when its vertical velocity is 0), vi= 69.6 m/s, and a=g=9.8 m/s^2, we get h= 602.3 m.

Since the maximum height is greater than 200 m, the boulder will violate the safety rule and not be able to clear the wall. To determine the height at which the boulder will hit the wall, we can use the equation x=(v0^2sin2theta)/g, where x is the horizontal distance traveled by the boulder. Plugging in the values of x=325 m (since the wall is 325 m away from the catapult), vi= 69.6 m/s, and theta=30 degrees, we get the height of impact to be 29.5 m. This means that the boulder will hit the wall at a height of 29.5 m, 7.5 m below the top of the wall.

In conclusion, the catapult is not suitable for the scene as it violates the safety rule and will hit the wall at a height of 29.5 m. To make the scene realistic and safe, alternative methods of
 

1. What is projectile motion?

Projectile motion is the motion of an object that is launched into the air and then moves under the influence of gravity. It follows a curved path known as a parabola.

2. What factors affect projectile motion?

The factors that affect projectile motion include the initial velocity, angle of launch, air resistance, and gravity. These factors can impact the trajectory, distance, and time of flight of the object.

3. How is projectile motion calculated?

Projectile motion is calculated using the equations of motion for both horizontal and vertical motion. The initial velocity, angle of launch, and acceleration due to gravity are used to determine the displacement, velocity, and time of flight of the object.

4. What is the difference between horizontal and vertical motion in projectile motion?

Horizontal motion refers to the motion of the object along the x-axis, while vertical motion refers to the motion along the y-axis. In projectile motion, the horizontal velocity remains constant, while the vertical velocity changes due to the influence of gravity.

5. What are some real-life examples of projectile motion?

Some real-life examples of projectile motion include throwing a ball, kicking a soccer ball, shooting a basketball, and launching a rocket. All of these objects follow a parabolic path due to the influence of gravity.

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