Question

A water balloon slingshot launches its projectiles essentially from ground level at a speed of 25.5 m/s . (You can ignore air resistance.)
A: At what angle should the slingshot be aimed to achieve its maximum range?

B: If shot at the angle you calculated in Part A, how far will a water balloon travel horizontally?

C: For how long will the balloon be in the air?

Answer 1

A) The longest horizontal distance is reached at 45 degrees angle. This is true for any projectile launch.

B) First, calculate fligth time (using the vertical motion) and then calculate the horizontal movement.

Flight time = 2* ascendent time

ascendent time => final vertical velocity, Vy, = 0

sin(45) = Voy / Vo => Voy = Vosin(45) = 25.5 m/s * (√2) / 2 = 18.03 m/s

Vy = Voy - gt = 0  => Voy = gt = t = Voy / g

Use g = 10 m/s^ (it is an aproximation, because the actual value is about 9.81 m/s^2 depending on the latitud)

t = 18.03 m/s / 10 m/s^2 = 1.83 s

This is the ascendant time going upward.

The flight time is 2*1.83 = 3.66 s

Horizontal motion

Horizontal velocity = Vx = constant = Vox = Vo*cos(45) = 18.03 m/s

Vx = x / t => x = Vx*t

Horizontal distance = xmax = 18.03m/s*3.66 s = 65.99 m

c) The time the ballon will be in the air was calculated in the part B, it is 18.03 s