(Velocity and acceleration.) The acceleration of gravity on the surface of the Moon is a = 1.6 m/s2, as opposed to g = 9.8 m/s2 on Earth. An Olympic high-jumper can jump to about 2.35 m on the surface of the Earth. What height could an Olympic high-jumper reach on the surface of the Moon?
m g h = (1/2) m v^2
h = v^2/2g
v the same
v^2 = 2 g h
2 * 9 .8 * 2.35 = 2 * 1.6 * h
h = 14.4 meters
To find the height that an Olympic high-jumper could reach on the surface of the Moon, we can use the following formula:
h = (v^2 - u^2) / (2*a)
Where:
h is the height reached by the high-jumper,
v is the final velocity of the high-jumper,
u is the initial velocity of the high-jumper,
a is the acceleration due to gravity.
On the surface of the Moon, the acceleration due to gravity (a) is given as 1.6 m/s^2, and the high-jumper's initial velocity (u) is 0 m/s since they start from rest. We need to find the final velocity (v) when the high-jumper reaches a height of 2.35 m.
Using the same formula, but rearranged, we have:
v^2 = u^2 + 2*a*h
Substituting the known values:
v^2 = 0 + 2*1.6*2.35
v^2 = 7.392
Taking the square root of both sides gives:
v ≈ 2.72 m/s
Therefore, the final velocity of the high-jumper is approximately 2.72 m/s. To find the height (h) that they could reach on the surface of the Moon, we can use the first formula:
h = (v^2 - u^2) / (2*a)
Substituting the known values:
h = (2.72^2 - 0^2) / (2*1.6)
h = 7.016 m
So, an Olympic high-jumper could reach a height of approximately 7.016 meters on the surface of the Moon.