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?
1. u = 25 cos 13 forever
2. Vi = 25 sin 13 up at start
3. top means zero velocity up
4. (1/2) m Vi^2 = m g h
h = Vi^2/(2g)
5. v = Vi - g t
0 = Vi - 9.81 t
t = Vi/9.81
6. same amount of time it took to go up
7. 2 t
8. - Vi
9. same old u
10. R = u (2 t) where t is the time up or the time down
To answer these questions, we will use the equations of motion for projectile motion. We can start by breaking down the initial velocity of the cannonball into its x and y components.
1. The x component of the cannonball's initial velocity, Voy, can be found using the equation:
Vox = V * cos(θ)
where V is the magnitude of the initial velocity (25 m/s) and θ is the launch angle (13°). Plugging in the values, we get:
Vox = 25 m/s * cos(13°) ≈ 24.36 m/s
2. The y component of the cannonball's initial velocity, Voy, can be found using the equation:
Voy = V * sin(θ)
Plugging in the values, we get:
Voy = 25 m/s * sin(13°) ≈ 5.57 m/s
3. At the peak of the cannonball's trajectory, the y component of the velocity (Vy) will be zero. This is because the projectile momentarily stops moving vertically before reversing direction. So, Vy = 0.
4. To find the height the cannonball reaches, we can use the formula:
V^2y = V^2oy + 2ay * Δy
Since the cannonball reaches its maximum height when Vy = 0, V^2y = 0. Therefore, we can simplify the formula to:
0 = V^2oy + 2ay * Δy
Rearranging the formula, we get:
Δy = -V^2oy / (2 * ay)
where ay is the acceleration due to gravity (-9.8 m/s^2). Plugging in the values, we get:
Δy = -(5.57 m/s)^2 / (2 * -9.8 m/s^2) ≈ 1.64 m
5. The time it takes for the cannonball to reach its peak can be found using the equation:
Vy = Voy + at
Plugging in the values, we have:
0 = 5.57 m/s + (-9.8 m/s^2) * t (since the initial velocity in the y direction is positive and acceleration is negative due to gravity)
Solving for t, we get:
t = 5.57 m/s / 9.8 m/s^2 ≈ 0.57 s
6. The time it takes for the cannonball to fall back down to the ground can be found using the same equation as in question 5.
Since the time it takes to rise to the peak is the same as the time it takes to fall from the peak to the ground, the time will also be 0.57 s.
7. The total time the cannonball spends in the air can be calculated by doubling the time it took to reach the peak.
Total time = 2 * 0.57 s = 1.14 s
8. The y component of the final velocity of the cannonball right before it hits the ground again can be found using the equation:
Vy = Voy + at
Plugging in the values, we have:
Vy = 5.57 m/s + (-9.8 m/s^2) * 0.57 s ≈ -0.59 m/s (negative because the final velocity is in the downward direction)
9. The x component of the final velocity of the cannonball right before it hits the ground again will be the same as the initial x component (Vox) since there is no horizontal acceleration.
10. The range (horizontal distance traveled) of the cannonball can be calculated using the equation:
Range = Vox * total time
Plugging in the values, we have:
Range = 24.36 m/s * 1.14 s ≈ 27.76 m
So, the answers to the questions are as follows:
1. Vox ≈ 24.36 m/s
2. Voy ≈ 5.57 m/s
3. Vy = 0 m/s
4. The cannonball reaches a height of approximately 1.64 m.
5. It takes approximately 0.57 s for the cannonball to reach its peak.
6. It takes approximately 0.57 s for the cannonball to fall back down to the ground.
7. The total time the cannonball spends in the air is approximately 1.14 s.
8. Vy ≈ -0.59 m/s
9. Vox ≈ 24.36 m/s
10. The range of the cannonball is approximately 27.76 m.