A ball is thrown with the speed of 10 m/s horizontally from the window of the 22^nd floor of one of two buildings and caught at the 8^th floor window. Each floor of the twin buildings is 5 meters high. What is the distance between the twin buildings?

Question

Quizplus · Accepted Answer

This problem provides the vertical distance (height of each floor and the difference between floors), vertical acceleration (gravitational constant), and the initial horizontal velocity (speed of the ball). Vx = Horizontal velocity (10 m/s) d_y = vertical displacement of the object d_x = unknown (this is what we are solving for) t = unknown (flight time) a = gravitational constant (9.8 m/s²) Since we assume there is no air resistance, d_h = V_xt. We solve this problem in three steps. Step 1: Calculate the vertical distance Vertical distance = (Floor difference)(floor height) Vertical distance = (22-8)(5 m) Vertical distance = 70 m Step 2: Calculate flight time Flight Time = √((vertical distance)/(1/2)(vertical acceleration)) Flight Time = √((70 m)/(1/2)(9.8 m/s²)) Flight Time = 3.78 s Step 3: Use the horizontal distance equation to solve for horizontal distance dx = v_xt dx = 10 m/s x 3.78 s dx = 37.8 m Since there is no air resistance, the ball will travel 37.8 m between buildings.

A Ball Is Thrown with the Speed of 10 M/s