A water balloon cannon is fired at 30m/s at an angle of 50∘ above the horizontal. How far away will it fall?

Answer 1

Here is a drawing:

Using trig ratios, cos(50°) = v_x/30 so v_x = 30cos(50°) = 19.2836 m/s

Also using trig ratios, sin(50°) = v_y/30 so v_y = 30sin(50°) = 22.9813 m/s

Considering ONLY the y-direction (up/down), the balloon acts under the influence of gravity. The kinematic equation of motion say v_f = v_i + at where v_f is the final velocity, v_i is the initial velocity, "a" is the acceleration (in this case a = g = -9.81 m/s²), and "t" is the time. At the top of the balloon's travel the velocity in the y-direction is zero. It travels with an initial velocity of 22.9813 and begins to slow, in the y-direction, as it rises until it reaches the top of it's arc of travel, then begins to fall, or have a negative velocity, speeding up, in the y-direction, until it hits the ground.

So, in the y-direction 0 = 22.9813 + (-9.81)t

9.81t = 22.9813

t = 2.3426 seconds. It then falls (in the y-direction) back to earth taking the same time t = 2.3426 seconds. So the total travel time is 2.3426•2 = 4.685 seconds

During that time, considering ONLY the x-direction, the balloon travels at 19.2836 m/s. The kinematic equation of motion when there is no acceleration (the acceleration only occurs in the y-direction, not in the x-direction) is d = vt where "d" is distance, "v" is velocity, and "t" is time.

So, d = 19.2836•4.685 = 90 meters