Practice!

Sample Problem

You throw an orange out a window at a height of 12.00 meters upwards at an angle of 32° to the horizontal at a velocity of 3.5 m/s. What is the maximum height that the orange will reach? Round the answer to the nearest hundredth.

meters

Solution

The maximum height is the height at time v0/g, where v0 in this case is the vertical component of the velocity vector. We know that v0/ (-g) = 3.5sin32/ (+ 9.81) = .189s. We plug this value into the equation for a falling object, .5gt2 + v0 t + h = x, and we get x = -4.9 (.189) 2 + (3.5sin32) .189 + 12 = 12.18