A playground is on the flat roof of a city school, 6.3 m above the street below

Question

A playground is on the flat roof of a city school, 6.3 m above the street below (see figure). The vertical wall of the building is h = 7.80 m high, forming a 1.5-m-high railing around the playground. A ball has fallen to the street below, and a passerby returns it by launching it at an angle of theta = 53.0 degree 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. (a) Find the speed at which the ball was launched. (Give your answer to two decimal places to reduce rounding errors in later parts.) 18.13m/s (b) Find the vertical distance by which the ball clears the wall 0.34 m (c) Find the horizontal distance from the wall to the point on the roof where the ball lands.

Explanation / Answer

Notice that every projectile formula contains 3 variables (actually 4, but acceleration is considered constant in projectile problems). You just have to use the formula where you can fit two of the given values and find the remaining variable. In this problem, you have the initial velocity with the angle, most of the distances, and the time at the top of the wall.

a) The ball takes 2.20 s to travel 24.0 m horizontally.
horizontal speed = 24.0/2.20 = 10.9 m/s
That's the horizontal component of the launch velocity, which was at a 53.0

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