Question

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?

Answer 1

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.