Downstream, the current is helping the swimmer go faster relative to the land
Answer:
25m/s²
Explanation:
Using one of the equations of motion.
v² = u²+2as where
v is the final velocity of the astronaut
u is his initial velocity
a = -g (the acceleration will be acceleration due to gravity since he is acting under the influence of gravity. The value is negative because the astronaut jumps up to a particular height)
s = H = total height covered
The equation will then become;
v² = u²-2gH
Given
u = 60m/s
v = 0m/s
g = ?
H = 72m
Substituting the given value into the equation;
0² = 60²-2g(72)
0 = 3600-144g
-3600 = -144g
g = -3600/-144
g = 25m/s²
The magnitude of his acceleration due to gravity on the planet is 25m/s²
The 4 fourth row of the graph is a quadratic graph
