Abaseball is thrown straight upward on the moon with an initial speed of 35 m/s.
compute:
(a) the maximum height reached by the ball. > (b) the time taken to reach that height.
--> (c) its velocity 30 s after it is thrown.
(d) when the ball's height is 100 m.

a) 376 m

b) 21.5 s

c) -13.9 m/s

d) 3.08 s, 39.9 s

Explanation:

Given:

y₀ = 0 m

v₀ = 35 m/s

a = -1.63 m/s²

a) Find: y when v = 0 m/s.

v² = v₀² + 2a (y − y₀)

(0 m/s)² = (35 m/s)² + 2(-1.63 m/s²) (y − 0 m)

y = 376 m

b) Find: t when v = 0 m/s.

v = at + v₀

0 m/s = (-1.63 m/s²) t + 35 m/s

t = 21.5 s

c) Find: v when t = 30 s.

v = at + v₀

v = (-1.63 m/s²) (30 s) + 35 m/s

v = -13.9 m/s

d) Find: t when y = 100 m.

y = y₀ + v₀ t + ½ at²

100 m = 0 m + (35 m/s) t + ½ (-1.63 m/s²) t²

100 = 35t − 0.815 t²

Solve with quadratic formula:

t = 3.08 s, 39.9 s

the answer is a. the yacht traveled from the dock north to the 200-meter marker in the bay in less than 5 minutes, giving the passengers several more hours to fish.

im doing peak rn so sry if its wrong