Answer :
By analyzing both vertical and horizontal motions, we determine that the initial horizontal speed of the ball is approximately 6.98 m/s.
To determine the initial speed of the ball thrown horizontally from the dorm window, we need to analyze its horizontal and vertical motion separately. The ball is initially at a height of 16.0 m and caught at a height of 1.55 m, 12.0 m away horizontally. We use the following steps:
Vertical Motion Analysis
The vertical displacement (
Δy) is the difference in height:
Δy = 16.0 m - 1.55 m = 14.45 m
We use the second kinematic equation for vertical motion:
Δy = vi,y * t + (1/2) * g * t²
Since the initial vertical velocity (vi,y) is 0 (as the ball is thrown horizontally), the equation simplifies to:
14.45 m = (1/2) * 9.81 m/s² * t²
Solving for t:
t² = 2 * 14.45 m / 9.81 m/s² ≈ 2.947
t = √2.947 ≈ 1.72 s
Horizontal Motion Analysis
As there is no acceleration in the horizontal direction, the horizontal distance (Δx) covered is:
Δx = vi,x * t
Given Δx = 12.0 m and t = 1.72 s:
vi,x = Δx / t = 12.0 m / 1.72 s ≈ 6.98 m/s
Thus, the initial speed of the ball is approximately 6.98 m/s.