A wrench falls out of the gondola of a balloon that is 520 m above the ocean.
Assuming negligible air resistance, what was the wrench’s speed as it hit the water? The acceleration of gravity is 9.8 m/s2 .
vf^2=2gh solve for Vf
vf^2=2*9.8*520
vf=100.955 ??
To find the wrench's speed as it hits the water, we can use the kinematic equation for free fall. The equation is:
v^2 = u^2 + 2as
where,
v = final velocity (which in this case is the wrench's speed as it hits the water)
u = initial velocity (in this case, the wrench is dropped from rest, so the initial velocity is 0)
a = acceleration due to gravity (given as 9.8 m/s^2)
s = displacement (in this case, the distance the wrench falls, which is 520 m)
Plugging in the given values into the equation, we get:
v^2 = 0^2 + 2 * 9.8 * 520
v^2 = 0 + 10192
v^2 = 10192
To find v, we take the square root of both sides:
v = √10192
v ≈ 101.9 m/s
Therefore, the wrench's speed as it hits the water is approximately 101.9 m/s.