A ball rolls horizontally off the edge of a cliff at 4.00 m/s. If the ball lands a distance of 30.0 m from the base of the vertical cliff, what is the height of the cliff?
A)276 m
B)138 m
C)9.20 m
D)552 m
E)92.0 m

Question

Quizplus · Accepted Answer

The correct answer is A because the ball has no vertical velocity when it leaves the cliff, so its vertical motion is determined by the acceleration due to gravity. The horizontal motion of the ball is not affected by gravity, so it moves at a constant velocity of 4.00 m/s. The time it takes for the ball to fall to the ground can be found using the equation:$$y = -0.5 * g * t^2$$where y is the height of the cliff, g is the acceleration due to gravity (9.8 m/s^2), and t is the time it takes for the ball to fall. The horizontal distance the ball travels during this time is:$$x = v * t$$where x is the horizontal distance, v is the horizontal velocity (4.00 m/s), and t is the time it takes for the ball to fall. Substituting the first equation into the second equation, we get:$$x = v * (-0.5 * g * y / v)^2$$Solving for y, we get:$$y = x^2 / (0.5 * g * v^2)$$Plugging in the given values, we get:$$y = 30.0^2 / (0.5 * 9.8 * 4.00^2) = 276 m$$

A Ball Rolls Horizontally Off the Edge of a Cliff