A bored kid, 30 meters from a building throws a ball at an angle of 50 degrees with a velocity of 20m/s at the building wall. at what height above the throwing line will the ball hit the wall?

9.1 meters First, determine the vertical and horizontal components of the ball's velocity. So Vhorz = cos (50) * 20 m/s = 0.64278761 * 20 m/s = 12.85575219 m/s Vvert = sin (50) * 20 m/s = 0.766044443 * 20 m/s = 15.32088886 m/s Now determine how long until the ball hits the building. Do this by dividing the horizontal distance by the horizontal velocity. T = 30 m / 12.85575219 m/s = 2.33358574 s Now the expression for the height of the ball is given by h = VT - 0.5 AT^2 where h = height V = initial vertical velocity T = Time A = acceleration due to gravity. Substituting known values. h = (15.32088886 m/s) (2.33358574 s) - 0.5 * 9.8 m/s^2 (2.33358574 s) ^2 h = 35.75260778 m - 4.9 m/s^2 * 5.445622407 s^2 h = 35.75260778 m - 26.6835498 m h = 9.069057982 m Rounding to 2 significant figures gives 9.1 meters. So the ball will impact the building 9.1 meters above the height at which the ball was thrown.