5. A ball rolls off a 1.5 m tall horizontal table and lands on the floor 0.70 m away. A. How much time is the ball in the air?
B. How does that time compare with the time it takes for a dropped ball to fall that same distance.
C. What is the ball's velocity while it was on the table top?
D. What is the horizontal component of its velocity just prior to impact?
E. What is the vertical component of its velocity just prior to impact?
F. What is the magnitude of its velocity just prior to impact?
G. What is the direction of its velocity just prior to impact?

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.

this means that the temperature of both substances will be equal