**50. In the annual battle of the dorms, students gather on the roofs of Jackson and Walton dorms to launch water balloons at each other with slingshots. The horizontal distance between the buildings is 35.0 m, and the heights of the Jackson and Walton buildings are, respectively, 15.0 m and 22.0 m. Ignore air resistance. (a) The first balloon launched by the Jackson team hits Walton dorm 2.0 s after launch, striking it halfway between the ground and the roof. Find the direction of the balloon’s initial velocity. Give your answer as an angle measured above the horizontal. (b) A second balloon launched at the same angle hits the edge of Walton’s roof. Find the initial speed of this second balloon.

The basic kinematic equation for both horizontal and vertical motion is

 

For horizontal motion this becomes:

Since there is no air resistance

For vertical motion this becomes

 

 

Solve both for the velocity – trig function terms as follows

 

 

and

 

Now divide these two equations and we get

 

 

 

 

Plug in values,  x is distance between dorms 35.0 m, y is difference between launch height and where it hits.  Launch height is 15.0 m and landing height is 11.0 m, so y = -4.0 m.  Finally we are told the time for the projectile motion is 2.0 s.

 

 

 

For second launch, y2 is +7.0 m launch height is still 15 m but landing height is now 22 m.  x is still 35 m.  Angle is still 24°.  This time we do not know the time of flight or the initial velocity which we are asked to find.  Using horizontal distance we can solve for time.

 

 

Now set up for y2

 

 

 

We need to solve for

 

 

 

 

Now we can plug in values

 

 

 

 

 

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This page last updated on January 11, 2020