an object of mass 10kg is dropped from a 20 m cliff. what is the velocity just before the object strikes the ground?
v = Vo + a t = -9.8 t
z = Zo + Vo t + (1/2) a t^2
0 = 20 + 0 + 4.9 t^2
so
t = sqrt (20/4.9)
and
v = -9.8 sqrt (20.4.9)
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To find the velocity just before the object strikes the ground, we can use the laws of motion and the concept of gravitational acceleration.
1. Determine the height of the cliff:
The height of the cliff is given as 20 m.
2. Calculate the final velocity using the formula:
Final velocity (v) = √(2 * acceleration due to gravity * height)
Acceleration due to gravity is typically approximated as 9.8 m/s^2 on Earth.
v = √(2 * 9.8 m/s^2 * 20 m)
v = √(196 m^2/s^2)
v ≈ 14 m/s
Therefore, the velocity just before the object strikes the ground is approximately 14 m/s.