posted by Caitlin on .
A ball is thrown from a point 1.38 m above the ground. The initial velocity is 19.9 m/s at an angle of 33.8° above the horizontal. Find the maximum height of the ball above the ground.
Calculate the speed of the ball at the highest point in the trajectory.
looking at energy, at the highest point, the ball will have no vertical KE, so total energy will be initial PE+initialKE
1/2 mVhoriz^2+mg(height)=1/2 m19.9^2+mg(1.38)
solve for height.
and Vhoriz is the speed at the hightest point.