Motion in 2D

Children playing in a playground on the flat roof of a city school lose their ball to the parking lot below. One of the teachers kicks the ball back up to the children as shown in the figure below. The playground is 5.40 m above the parking lot, and the school building's vertical wall is h = 6.60 m high, forming a 1.20 m high railing around the playground. The ball is launched at an angle of 9 = 53.0° above the horizontal at a point d = 24.0 m from the base of the building wall. The ball takes 2.20 s to reach a point vertically above the wall. (Due to the nature of this problem, do not use rounded intermediate values in your calculations—including answers submitted in WebAssign.) (a) Find the speed (in m/s) at which the ball was launched. m/s (b) Find the vertical distance (in m) by which the ball clears the wall. m (c) Find the horizontal distance (in m) from the wall to the point on the roof where the ball lands. m (d) What If? If the teacher always launches the ball with the speed found in part (a), what is the minimum angle (in degrees above the horizontal) at which he can launch the ball and still clear the playground railing? (Hint: You may need to use the trigonometric identity sec2(9) = 1 + tan2(9).) ° above the horizontal (e) What would be the horizontal distance (in m) from the wall to the point on the roof where the ball lands in this case? m