I believe it is, All of the above.
Answer:

Explanation:
At some distance from the Earth the force of attraction due to moon is balanced by the force due to Moon
so we will have

now we have


so we will have

Now by energy conservation



Answer:
3.467 s
Explanation:
given,
distance , d = 49 mm = 0.049 m
initial speed of the of the rock, v = 17 m/s
time taken by the Heather rock to reach water
using equation of motion

taking downward as negative

4.9 t² + 17 t - 0.049 = 0
now,
t₁ = -3.47 s , 0.0028 s
rejecting negative values
t₁ = 0.0028 s
now, time taken by the ball of Jerry
using equation of motion

taking downward as negative

4.9 t² - 17 t - 0.049 = 0
now,

t₂ = 3.47 s ,-0.0028 s
rejecting negative values
t₂ = 3.47 s
now, time elapsed is = t₂ - t₁ = 3.47 - 0.0028 = 3.467 s
It would die from the lack of nourishment