Take the starting position of the ball 1.5 m above the floor to be the origin. Then at time t, the ball's horizontal and vertical positions from the origin are
x = v₀ t
y = -1/2 gt²
where v₀ is the initial speed with which it rolls off the edge and g = 9.8 m/s².
A. The floor is 1.5 m below the origin, so we solve for t when y = -1.5 m :
-1.5 m = -1/2 gt²
⇒ t² = (3.0 m)/g
⇒ t = √((3.0 m)/g) ≈ 0.55 s
B. It would take the same amount of time.
C. The ball travels a horizontal distance of 0.70 m before reaching the floor, so we solve for v₀ with t = 0.55 s :
0.70 m = v₀ (0.55 s)
⇒ v₀ = (0.70 m) / (0.55 s) ≈ 1.3 m/s
D. At time t, the ball has horizontal and vertical velocity components
v[x] = 1.3 m/s
v[y] = -gt
so the horizontal component of the ball's final velocity vector is the same as the initial one, 1.3 m/s.
E. The vertical component of velocity would be
v[y] = -g (0.55 s) ≈ -5.4 m/s
F. The magnitude of the final velocity would be
√((1.3 m/s)² + (-5.4 m/s)²) ≈ 5.6 m/s
G. The final velocity vector makes an angle θ with the horizontal such that
tan(θ) = (-5.4 m/s) / (1.3 m/s)
⇒ θ = arctan(-5.4/1.3) ≈ -77°
i.e. approximately 77° below the horizontal.
