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

The maximum height is the height at time v_{0}/g,
where v_{0} in this case is the vertical component of the velocity
vector. We know that v_{0}/ (-g) = 3.5sin32/ (+ 9.81) = .189s. We plug
this value into the equation for a falling object, .5gt^{2} + v_{0}
t + h = x, and we get x = -4.9 (.189) ^{2} + (3.5sin32) .189 + 12 =
12.18