From rest, a rock is dropped from a garage roof. The roof is 6.0 meters above ground level. The rock will reach the earth at a speed of 10.849 meters per second.
The change of displacement with respect to time is defined as the velocity. Velocity is a vector quantity.
it is a time-based component. Velocity at any angle is resolved to get its component of x and y-direction.
Given data:
V(Final velocity)=? (m/sec)
h(height)= 6.0 m
u(Initial velocity)=0 m/sec
g(gravitational acceleration)=9.81 m/s²
Newton's third equation of motion:
[tex]\rm v_y^2 = u_y^2+ 2gh \\\\\rm v_y^2 = 0+ 2gh\\\\\ v_y= \sqrt{2\times 9.81 \ (m/s^2)\times 6.0 (m)} \\\\ v_y=10.849 \ m/sec[/tex]
Hence, the velocity of the rock as it hits the ground will be 10.849 m/sec.
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