# Bob was dropped from a cliff and hit the ground with a speed of 90m/s. What was the height of the cliff assuming acceleration due to gravity is -9.8m/s^2?

## Expert Answers

Hello!

I think we ignore air resistance. In that case, free fall is a uniformly accelerated motion. The constant acceleration is `g=9.8 m/s^2.`

Therefore the speed changes with time as `V(t)=V_0+g t,` where `V_0` is the initial speed (downwards). In our problem the initial speed is probably zero, so `V=g...

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Hello!

I think we ignore air resistance. In that case, free fall is a uniformly accelerated motion. The constant acceleration is `g=9.8 m/s^2.`

Therefore the speed changes with time as `V(t)=V_0+g t,` where `V_0` is the initial speed (downwards). In our problem the initial speed is probably zero, so `V=g t.` Denote the final speed as `V_1,` then the time of the falling was `t_1=V_1/(g) = 90/9.8 approx 9.2 (s).`

The formula for the height is `H(t)=H_0-V_0 t -(g t^2)/2,` where `H_0` is the initial height. So `H(t_1)=0=H_0-(g t_1^2)/2 = V_1^2/(2g)` and from this we obtain `H_0=(V_1^2)/(2g) approx 413 (m).` This is the answer.

Another approach uses energy conservation: `mgH_0=(mV_1^2)/2,` so `H_0=(V_1^2)/(2g)` also.

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