Question

What would be the speed of an object just before hitting the ground if dropped 91.5 meters

Answer 1

about 42.35 m/s

Step-by-step explanation:

Use the equation for accelerated motion (g), and with zero initial velocity that doesn't include time:

`v_f^2=v_i^2+2\,a\,\Delta x`

which for our case would reduce to:

`v_f^2=v_i^2+2\,a\,\Delta x\\v_f^2=0+2\,9.8\,(91.5)\\v_f^2= 1793.4\\v_f=√(1793.4) \\v_f \approx 42.35`

then the velocity just before hitting would be about 42.35 m/s