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?

1 Answer
Apr 24, 2018

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.