A cannon shoots a cannonball with an initial speed of 25 m/s at an angle 13.
1. What is the x component of the cannonballs initial velocity Vox?
2. What is the y component of the cannonballs initial velocity Voy?
3. What will be Vy, the y component of the cannonballs final velocity at its peak? Refer to problem a
4. How high does the cannonball go? V^2y=V^2oy+2ay(delta y)
5. How long will it take the cannonball to reach it's peak?
6. How long will it take the cannonball to fall back down to the ground again?
7. What will be the total time the cannonball spends in the air?
8. What will be the y component of the final velocity of the cannonball right before it hits the ground again?
9. What will be the x component of the final velocity of the cannonball right before it hits the ground again?
10. What will be the range (horizontal distance traveled) of the cannonball?
The procedure is the same as your 8:56 PM. post.
To solve these problems, we will use the equations of projectile motion. The initial velocity of the cannonball is given as 25 m/s at an angle of 13°.
1. The x component of the initial velocity, Vox, can be found using the equation Vox = V0 * cos(θ), where V0 is the initial velocity and θ is the angle. Substituting the values given, we get: Vox = 25 m/s * cos(13°). Calculate the value to get the answer.
2. The y component of the initial velocity, Voy, can be found using the equation Voy = V0 * sin(θ), where V0 is the initial velocity and θ is the angle. Substituting the values given, we get: Voy = 25 m/s * sin(13°). Calculate the value to get the answer.
3. At the peak of its trajectory, the vertical component of velocity, Vy, becomes zero. Therefore, the answer to this question will be zero.
4. To find the maximum height reached by the cannonball, we can use the equation V^2y = V^2oy + 2ay(delta y), where V^2y is the final velocity (zero at the peak), V^2oy is the initial vertical velocity, ay is the acceleration due to gravity (-9.8 m/s^2), and delta y is the change in height (unknown). Rearranging the equation, we get delta y = V^2oy / (2 * ay). Substituting the values given, we have: delta y = (25 m/s * sin(13°))^2 / (2 * (-9.8 m/s^2)). Calculate the value to get the answer.
5. To find the time it takes for the cannonball to reach its peak, we can use the equation Vy = Voy + ay * t, where Vy is the final vertical velocity (zero at the peak), Voy is the initial vertical velocity, ay is the acceleration due to gravity, and t is the time. Substituting the values given, we get: 0 = 25 m/s * sin(13°) + (-9.8 m/s^2) * t. Rearrange the equation to solve for t: t = -25 m/s * sin(13°) / (-9.8 m/s^2). Calculate the value to get the answer.
6. The time it takes for the cannonball to fall back down to the ground is the same as the time it took to reach the peak. So the answer will be the same as the answer for question 5.
7. The total time the cannonball spends in the air can be found by multiplying the time to reach the peak by 2, since the time to go up and come back down is the same. So the answer will be 2 times the answer for question 5.
8. The y component of the final velocity of the cannonball right before it hits the ground is the same as the initial vertical velocity, but with opposite sign. So the answer will be -Voy.
9. The x component of the final velocity of the cannonball right before it hits the ground remains the same as the initial x component, Vox. So the answer will be the same as the answer for question 1.
10. The range (horizontal distance traveled) of the cannonball can be found using the equation Range = Vox * t, where Vox is the x component of the initial velocity and t is the total time the cannonball spends in the air (answer to question 7). Multiply the values to get the answer.
Remember to perform the necessary calculations and use appropriate units to get the final answers.