Question

A ball is thrown so that its initial vertical and horizontal components of velocity are 60 m/s and 15 m/s, respectively. Use a motion diagram to estimate the ball's total time of flight and the distance it traverses before hitting the ground.

Answer 1

The ball is in the air for approximately 6.12 seconds and traverses a horizontal distance of about 91.8 meters before hitting the ground.

Step-by-step explanation:

In order to estimate the total time of flight and distance traversed by the ball before hitting the ground, we can use the concept of projectile motion. Considering the horizontal and vertical components of velocity, we can break down the motion of the ball.

First, let's calculate the time of flight. Since the vertical motion is influenced by gravity, we can use the equation d = V*t + (1/2)*g*t^2, where d is the displacement in the vertical direction, V is the initial vertical velocity, t is the time of flight, and g is the acceleration due to gravity. In this case, the displacement is zero (the ball is vertically at the same height as where it was launched), so the equation becomes 0 = (60*t) + (1/2)(-9.8*t^2). Solving for t, we get t ≈ 6.12 s. Therefore, the ball is in the air for approximately 6.12 seconds.

Next, let's determine the distance traversed horizontally. Since there is no horizontal acceleration, the ball's horizontal velocity remains constant throughout its motion. Therefore, we can use the equation d = V*t, where d is the horizontal distance, V is the initial horizontal velocity, and t is the time of flight. Substituting the given values, we have d = 15 * 6.12 ≈ 91.8 m. Hence, the ball traverses a horizontal distance of approximately 91.8 meters before hitting the ground.