Question

A seaside cliff is 30 m above the ocean surface, and Sam is standing at the edge of the cliff. Sam has three identical stones. The first stone he throws off the cliff at 30° above the horizontal. The second stone he throws vertically downward into the ocean. The third stone he drops into the ocean.
1. In terms of magnitude, which stone has the largest change in its velocity over a one second time interval after its release? (Sam’s throwing speed is 10 m/s.)

Answer 1

Answer:

In terms of magnitude, the stones 2 and 3 have the largest change in its velocity over a one second time interval after their release.

Explanation:

Stone 1:

vi = 10 m/s

vix = vi*Cos ∅ = (10 m/s)*Cos 30° = 8.66 m/s = vx

viy = vi*Sin ∅ = (10 m/s)*Sin 30° = 5 m/s

vy = viy - g*t = (5 m/s) - (9.8m/s²)*(1 s) = -4.8

then

v = √(vx²+vy²) = √((8.66)²+(-4.8)²) = 9.90 m/s

Δv = v - vi = 9.902 m/s - 10 m/s

⇒  Δv = -0.098 m/s

Stone 2:

vi = 10 m/s

v = vi + g*t = (10 m/s) + (9.8m/s²)*(1 s) = 19.8 m/s

Δv = v - vi = (19.8 m/s) - (10 m/s)

⇒  Δv = 9.8 m/s

Stone 3:

vi = 0 m/s

v = g*t = (9.8m/s²)*(1 s) = 9.8 m/s

Δv = v - vi = (9.8 m/s) - (0 m/s)

⇒  Δv = 9.8 m/s

Finally, in terms of magnitude, the stones 2 and 3 have the largest change in its velocity over a one second time interval after their release.