Verified answer

Time of flight, T = 2vsinθ/g = 2(16.5)sin33 / 9.8 = 1.8 s

Horizontal component of velocity = 16.5 cos33 = 13.8 m/s

Horizontal distanced travelled by the catapult, R = v²sin2θ/g = 16.5²sin(2*33) / 9.8 = 25.4 m

To solve this problem, we must assume that the final height of the vat is same as its initial height. This means the time to rise from the initial height to its maximum height is equal to the time to fall from the maximum height to its initial height. The first steps are to determine the vertical and horizontal component of the initial velocity.

Vertical = 16.5 * sin 33

This is approximately 9 m/s.

Horizontal = 16.5 * cos 33

This is approximately 13.8 m/s. During one half of the total time, the vertical velocity decrease from approximately 9 m/s to 0 m/s at the rate of 9.8 m/s each second. Let’s use the following equation to determine the time for this to happen.

vf = vi + g * t

0 = 16.5 * sin 33 + -9.8 * t

t = 16.5 * sin 33 ÷ 9.8

Total time = 16.5 * sin 33 ÷ 4.9

The total time is approximately 1.83 seconds. To determine the horizontal distance the vat moves during this time, use the following equation.

d = v * cos θ * t

d = 16.5 * cos 33 * 16.5 * sin 33 ÷ 4.9 = 272.25 * cos 33 * sin 33 ÷ 4.9

This is approximately 25.4 meters. To check this answer, let’s use the following equation.

Range = v^2/g * sin 2θ

Range = 272.25/9.8 * sin 66

This is approximately 25.4 meters. I hope this is helpful for you.

What is the horizontal component of the velocity?

Answer in units of m/s. 16.5 * cos 33

How far from the catapult does it land? Answer in units of m.

272.25/9.8 * sin 66