(b) The initial velocity is added to that due to acceleration by gravity. The velocity is increased linearly by gravity at the rate of 9.8 m/s². The average velocity of the pebble will be its velocity halfway through the 2-second time period.* That is, it will be ...

4 m/s + (9.8 m/s²)(2 s)/2 = 13.8 m/s . . . . average velocity

(a) The distance covered in 2 seconds at an average velocity of 13.8 m/s is ...

d = vt

d = (13.8 m/s)(2 s) = 27.6 m

The water is about 27.6 m below ground.

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* We have chosen to make use of the fact that the velocity curve is linear, so the average velocity is half the sum of initial and final velocities:

vAvg = (vInit + vFinal)/2 = (vInit + (vInit +at))/2 = vInit +at/2

If you work this in a straightforward way, you would find distance as the integral of velocity, then find average velocity from the distance and time.

$\displaystyle d=\int_0^t{(v_0+at)}\,dt=v_0t+\dfrac{1}{2}at^2=t\left(v_0+a\dfrac{t}{2}\right)\\\\v_{avg}=\dfrac{d}{t}=v_0+a\dfrac{t}{2}\qquad\text{the formula we started with}$

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2 years ago