(a)

Physics

1 answer:

Lisa [10]2 years ago

4 0

(a) Let the base of the cliff be the origin, so that the rock's height y at time t is

y = h + (8.19 m/s) t - 1/2 gt²

where h is the height of the cliff. The rock hits the ground when y = 0. Solve for h when t = 2.37 s :

0 = h + (8.19 m/s) (2.37 s) - 1/2 (9.80 m/s²) (2.37 s)²

⇒ h ≈ 8.11 m

(b) If the rock is thrown straight down with the same speed, its height at time t would be

y = 8.11 m - (8.19 m/s) t - 1/2 gt²

Solve for t when y = 0 (use the quadratic formula and ignore the negative solution) :

t ≈ 0.699 s

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