# Motion Under Gravity

The height s (in feet) at the time t (in seconds) of a silver dollar dropped from the top of the Washington monument is

s=-16t+555

a.) Find the average velocity on the interval [2,3]

b.) FInd the instantaneous velocity when t=2 and when t=3

c.) how long will it take the dollar to hit the ground?

d.) Find the velocity of the dollar when it hits the ground.

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## SOLUTION This solution is **FREE** courtesy of BrainMass!

Sol.

In the expression of s the exponent of time is one hence the body is moving with uniform velocity.

s=-16t+555 ............(1)

gives velocity as a function of time t as

v = ds/dt = -16 + 0 = - 16 feet/second (independent of t) ...................(2)

a.) As the velocity is constant the average velocity is same as the instantaneous velocity for any interval of time. Hence the average velocity for interval [2,3] is - 16 feet/sec. or 16 feet/sec. downwards.

b.) Instantaneous velocity at t = 2 and at t = 3 is -16 feet/second.

c.) When the dollar hits the ground s = 0, thus eq (1) gives

0 = -16t + 555 or t = 555/16 =34.69 s

d.) velocity of the dollar when it hits the ground will be again -16 feet/second.

[ check your question I think it should be s = -16t^2 + 555 then instantaneous velocity at time t is

v = ds/dt = -16(2t) = -32t feet/sec then substitute the values of time toget instantaneous velocities.

For average velocity find s3 - s2 and divided by interval of time 1s.]

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