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.