A cannon fires a free projectile at an angle 85 from the ground. The cannonball has a speed of 150 m/s and it is launched from a hill 20 m above the plain where it will land
1. What is the x component of the cannonballs initial velocity
2. What is the y component of the cannonballs initial velocity
3. What is the x component of the cannonballs velocity at the peak of its trajectory?
4. What is the y component of the cannonballs velocity at the peak of its trajectory?
5. What is the cannonballs acceleration at the peak of its trajectory?
6. What is the height above the plain of the cannonball at the peak of its trajectory?
7. How long will the cannonball take to reach the peak of its trajectory?
8. How long will the cannonball take to fall from the ground from the peak?
9. What is the total time the cannonball will spend in the air?
10. What will be the y component of the final velocity of the cannonball when it hits the ground?
11. What is the total magnitude of the final velocity of the cannonball when it hits the ground?
12. What is the range of the cannonball?
13. At what angle will the cannonball hit the ground?
Hey, learn how to do the last one, then do this one. I took the course (in 1954) and do not need the practice.
To answer these questions, we can break down the motion of the cannonball into horizontal and vertical components.
1. The x component of the cannonball's initial velocity:
To find the x component, we use the equation: vx = v * cos(θ), where vx is the x component of velocity, v is the magnitude of the initial velocity, and θ is the launch angle. Given that the magnitude of the initial velocity is 150 m/s and the launch angle is 85°, we can calculate the x component of velocity: vx = 150 * cos(85°).
2. The y component of the cannonball's initial velocity:
To find the y component, we use the equation: vy = v * sin(θ), where vy is the y component of velocity, v is the magnitude of the initial velocity, and θ is the launch angle. Given that the magnitude of the initial velocity is 150 m/s and the launch angle is 85°, we can calculate the y component of velocity: vy = 150 * sin(85°).
3. The x component of the cannonball's velocity at the peak of its trajectory:
At the peak of the trajectory, the y component of the velocity becomes zero, so the x component remains constant. Therefore, the x component of velocity at the peak is the same as the initial x component: vx = 150 * cos(85°).
4. The y component of the cannonball's velocity at the peak of its trajectory:
At the peak of the trajectory, the y component of the velocity becomes zero. Therefore, the y component of velocity at the peak is zero.
5. The cannonball's acceleration at the peak of its trajectory:
The acceleration due to gravity acts vertically downward and is constant throughout the motion of the cannonball. Therefore, the acceleration at the peak of the trajectory is the same as the acceleration at any other point, which is approximately -9.8 m/s^2.
6. The height above the plain of the cannonball at the peak of its trajectory:
To find the height at the peak, we need to determine the maximum height reached by the cannonball. The maximum height can be found using the equation: y = (vy^2)/(2 * g), where y is the maximum height, vy is the y component of the initial velocity, and g is the acceleration due to gravity. Given that vy = 150 * sin(85°) and g is approximately 9.8 m/s^2, we can calculate the maximum height y.
7. The time it will take for the cannonball to reach the peak of its trajectory:
To find the time, we can use the equation: t = vy / g, where t is the time of flight to reach the peak, vy is the y component of the initial velocity, and g is the acceleration due to gravity. Given that vy = 150 * sin(85°) and g is approximately 9.8 m/s^2, we can calculate the time t.
8. The time it will take for the cannonball to fall from the peak to the ground:
The time it takes for the cannonball to fall from the peak to the ground is the same as the time to reach the peak. This is because the vertical motion is symmetrical. Therefore, the time to fall from the peak is also t.
9. The total time the cannonball will spend in the air:
To find the total time of flight, we can use the equation: T = 2 * t, where T is the total time of flight and t is the time to reach the peak. Given that t was calculated in the previous question, we can find T.
10. The y component of the final velocity of the cannonball when it hits the ground:
The y component of the final velocity when the cannonball hits the ground can be found using the equation: vy_final = -v * sin(θ), where vy_final is the y component of the final velocity, v is the magnitude of the initial velocity, and θ is the launch angle. Given that v = 150 m/s and θ = 85°, we can calculate vy_final.
11. The total magnitude of the final velocity of the cannonball when it hits the ground:
The total magnitude of the final velocity can be found using the equation: vf = sqrt(vx^2 + vy_final^2), where vf is the magnitude of the final velocity, vx is the x component of the initial velocity, and vy_final is the y component of the final velocity. Given that vx was calculated in question 1, and vy_final was calculated in question 10, we can calculate vf.
12. The range of the cannonball:
The range of the projectile is the horizontal distance it travels before hitting the ground. It can be calculated using the equation: R = vx * T, where R is the range, vx is the x component of the initial velocity, and T is the total time of flight. Given that vx was calculated in question 1, and T was calculated in question 9, we can calculate the range of the cannonball.
13. The angle at which the cannonball hits the ground:
The angle at which the cannonball hits the ground can be calculated using the equation: θ_final = atan(vy_final / vx), where θ_final is the angle of the final velocity, vy_final is the y component of the final velocity, and vx is the x component of the initial velocity. Given that vy_final was calculated in question 10, and vx was calculated in question 1, we can calculate θ_final.