Question

A water balloon with a mass of 500 g is launched across a waterpark and lands 150 m away 5 seconds later. What is the initial velocity in the x and y directions, launch angle, and max height?

Answer 1

The launch angle was approximately `39.2^@`. The vertical launch velocity was 24.5 m/s. The horizontal launch velocity was 30 m/s. The maximum height was 30.625 m.

Explanation:

Let

`theta` = the launch angle

`v` = the launch velocity

`g` = the acceleration of gravity = 9.8 `m/s^2`

Assume no air resistance and that the water balloon was launched at ground level.

Horizontal velocity = `vcostheta`

We know that the balloon traveled 150 meters horizontally in 5 seconds so

`vcostheta=150/5=30` m/s

The horizontal distance, `x`, as a function of time, `t`, is

`x= 30t`

which implies that

`t=x/30`.

The vertical height, `y`, as a function of time, `t`, is

`y=v(sintheta)t-(g)/2t^2`

Use the last two equations to eliminate `t`.

`y=v(sintheta)x/30-g/2x^2/900`

The balloon lands when `x=150`, so `y=0` when `x=150`.

`0=v(sintheta)150/30-g/2(150^2/900)`

`v(sintheta)=g/2(30/150)(150^2/900)=(9.8(150))/(2(30))=(9.8(5))/2=24.5` m/s

`v(sintheta)=24.5` m/s is the vertical launch velocity.

Notice that

`tantheta=(vsintheta)/(vcostheta)=24.5/30`

so

`theta=tan^-1(24.5/30)~~39.2^@`

The max height will be halfway through the flight or when `t=5/2`.

Remember that `y=v(sintheta)t-g/2t^2`, so the max height is

`y=24.5(5/2)-9.8/2(5/2)^2=30.625` m.