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Question

A person standing at the edge of a cliff throws a rock directly upward. it was observe that 2 seconds later the rock is at its maximum height (in ft) and that 5 seconds after that, it hits the ground at the base of the cliff.

a. what is the initial velocity of the rock?
b. how high is the cliff?
c. what is the velocity of the rock at time t?
d. with what velocity does the rock hit the ground?

Answer 1

Since the rock undergoes a constant downward acceleration of -9.8 m/s², we can compute:

v(t) = v_0 - a*t
s(t) = integral{0,t}[v(x)dx] = s_0 + v_0*t - ½at²

Since v(2) = 0 (the upward velocity stopped, and turned to downard velocity); we can compute:

0 = v_0 - 2*9.8 ==> v_0 = 19.6 m/s

Using the equation for s(t), but shifting the time reference for the beginning of the fall:

s(t') = s_0 + v_0*t' - ½a(t')² = 19.6 + 0 - 4.9*(t')²

At t' = 3 (3 seconds past the top of the arc),

s(t'=3) = 19.6 - 4.8*9 = -24.5 m. Since we took the 0 point of the height to be the top of the cliff, the cliff must be 24.5 m above the base.

The final velocity of the rock is given by v(t') = v_0 - a*t' = 0 - 9.8*3 = -29.4 m/s

So:

a. 19.6 m/s
b. 24.5 m
c. v(t) = (19.6 - 9.8*t) m/s
d. v_f = -29.4 m/s

Answer 2

We know we have an equation of the form s(t) = -16t^2 +bt + c. We also know that
(1) s(0) = height of the cliff = c,
(2) v(2) = 0, and
(3) s(7) = 0.

a. s'(t) = v(t) = -32t + b. From (2) we get -32(2) + b = 0, so b = 64. The initial velocity is v(0) = b = 64.

b. From (3) we find c = 336. We now know s(t) completely, so we find (from (1)) that s(0) = 336.

c. See a.

d. Find v(7).

Answer 3

the given data is insufficient