Question

Mr. Frey sets up his potato cannon on the roof of the High School, 10m off the ground. He angles it up 32.50 up from the horizontal. If the cannon fires at 16m/s, how far along the ground will the cannon hit from the edge of the building?

Answer 1

First, let's focus on the vertical component. The cannonball is shot upwards at an angle of 32.50° from the horizontal. Since the cannon is 10m off the ground, the initial vertical position of the cannonball is 10m.

Using the equation of motion for vertical motion, we can find the time it takes for the cannonball to reach its maximum height. The equation is:

v_f = v_i + at

Since the cannonball is shot upwards, the final vertical velocity (v_f) will be 0 m/s at its maximum height. The initial vertical velocity (v_i) is 16 m/s (since the cannon fires at that speed), and the acceleration (a) due to gravity is -9.8 m/s^2 (negative because it acts in the opposite direction of motion).

0 = 16 - 9.8t

Simplifying the equation, we find:

t = 16/9.8

t ≈ 1.63 seconds

Now that we know the time it takes for the cannonball to reach its maximum height, we can find the total time of flight. Since the cannonball will take the same amount of time to come down as it took to go up, the total time of flight is:

2t ≈ 2 × 1.63 ≈ 3.26 seconds

Next, let's focus on the horizontal component. The initial horizontal velocity (v_x) is the same as the cannonball's speed, which is 16 m/s. The horizontal distance traveled by the cannonball (d) can be calculated using the equation:

d = v_x × t

d = 16 × 3.26

d ≈ 52.16 meters

Therefore, the cannonball will hit the ground approximately 52.16 meters away from the edge of the building.