A cannonball is shot (from ground level) with an initial horizontal velocity of 38.0 m/s and an initial vertical velocity of 26.0 m/s.
1) What is the initial speed of the cannonball?
2) What is the initial angle θ of the cannonball with respect to the ground?
3) What is the maximum height the cannonball goes above the ground?
4) How far from where it was shot will the cannonball land?
5) What is the speed of the cannonball 2.8 seconds after it was shot?
6) How high above the ground is the cannonball 2.8 seconds after it is shot?
I will be happy to check your work.
I need to know how to do them first.
To answer these questions, we can use the basic principles of projectile motion. Here is how you can find the answers:
1) The initial speed of the cannonball can be found using the Pythagorean theorem. The horizontal and vertical velocities form the legs of a right triangle. So, the initial speed is calculated as follows:
Initial speed = √(horizontal velocity^2 + vertical velocity^2)
In this case, the horizontal velocity is 38.0 m/s and the vertical velocity is 26.0 m/s. Plugging these values into the formula, we get:
Initial speed = √(38.0^2 + 26.0^2)
Initial speed = √(1444 + 676)
Initial speed = √(2120)
Initial speed ≈ 45.99 m/s
Therefore, the initial speed of the cannonball is approximately 45.99 m/s.
2) The initial angle θ of the cannonball with respect to the ground can be found using trigonometry. We use the inverse tangent function (tan⁻¹) to find the angle. The equation for this is:
θ = tan⁻¹(vertical velocity / horizontal velocity)
Plugging in the values, we get:
θ = tan⁻¹(26.0 / 38.0)
θ ≈ 35.92 degrees
Therefore, the initial angle θ of the cannonball with respect to the ground is approximately 35.92 degrees.
3) The maximum height the cannonball goes above the ground can be determined by analyzing the vertical motion of the projectile. We will use the equation for vertical displacement:
Vertical displacement = (vertical velocity^2) / (2 * acceleration due to gravity)
The acceleration due to gravity is approximately 9.8 m/s². Plugging in the values, we get:
Vertical displacement = (26.0^2) / (2 * 9.8)
Vertical displacement ≈ 35.86 m
Therefore, the maximum height the cannonball goes above the ground is approximately 35.86 meters.
4) To find how far from where it was shot the cannonball will land, we need to analyze the horizontal motion. Since there is no horizontal acceleration, the horizontal velocity remains constant. We can use the formula:
Horizontal distance = horizontal velocity * time of flight
The time of flight is the total time it takes for the cannonball to reach the ground. We can find that using the vertical motion:
total time = (2 * vertical velocity) / (acceleration due to gravity)
Plugging in the values, we find:
total time = (2 * 26.0) / 9.8
total time ≈ 5.31 s
Now we can calculate the horizontal distance:
Horizontal distance = 38.0 * 5.31
Horizontal distance ≈ 201.78 m
Therefore, the cannonball will land approximately 201.78 meters from where it was shot.
5) To find the speed of the cannonball 2.8 seconds after it was shot, we need to calculate its velocities at that specific time. The horizontal velocity remains constant, so it will still be 38.0 m/s. The vertical velocity changes due to the acceleration of gravity.
Vertical velocity at a given time = initial vertical velocity - (acceleration due to gravity * time)
Plugging in the values, we get:
Vertical velocity at 2.8 s = 26.0 - (9.8 * 2.8)
Vertical velocity at 2.8 s ≈ 0.84 m/s
Now we can use the Pythagorean theorem to find the speed:
Speed at 2.8 s = √(horizontal velocity^2 + vertical velocity^2)
Speed at 2.8 s = √(38.0^2 + 0.84^2)
Speed at 2.8 s ≈ 38.01 m/s
Therefore, the speed of the cannonball 2.8 seconds after it was shot is approximately 38.01 m/s.
6) To find the height above the ground 2.8 seconds after the cannonball was shot, we can use the equation for vertical displacement:
Vertical displacement = (initial vertical velocity * time) + (0.5 * acceleration due to gravity * time^2)
Plugging in the values, we get:
Vertical displacement = (26.0 * 2.8) + (0.5 * 9.8 * 2.8^2)
Vertical displacement ≈ 47.63 m
Therefore, the cannonball is approximately 47.63 meters above the ground 2.8 seconds after it was shot.