Answer to Question #206817 in Mechanical Engineering for Rodel

Question

A golf ball is driven with a velocity of 60 m/s at an angle of 37° above the horizontal. It strikes a green at a horizontal distance of 250m from the tee. What was the elevation of the green above the tee?

Answer 1

elevation of the green above the tee $\frac{V_0^2(sin\theta)^2}{2g} = \frac{60^2\times sin30^2}{2\times9.81}=45.871m$