Question

A water balloon is launched at a speed of 34.0 m/s and an angle of 36 degrees above the horizontal. The water balloon hits a tall building located 23.0 m from the launch pad. At what height above the ground level will the water balloon hit the building? Calculate the answer in meters (m) and rounded to three significant figures.

Answer 1

Consider horizontal component of motion,

$v_{x} = \frac{x}{t}$

$t = \frac{x}{v_{x}}$

$t = \frac{23}{34 *cos 36}$

$t = 0.8362 s$

Now, consider vertical component of motion which is affected by acceleration due to gravity, in a downward direction

$s = ut +0.5 at^{2}$

$h = (34*sin36)*(0.8362)+0.5 *(-9.81)*0.8362^{2}$

$h = 13.2815 m$