A stone is dropped from the upper observation deck of a tower, 550 m above the g

Question

A stone is dropped from the upper observation deck of a tower, 550 m above the ground. (Assume g = 9.8 m/s2.) (a) Find the distance (in meters) of the stone above ground level at time t. h(t) = (b) How long does it take the stone to reach the ground? (Round your answer to two decimal places.) s (c) With what velocity does it strike the ground? (Round your answer to one decimal place.) m/s (d) If the stone is thrown downward with a speed of 9 m/s, how long does it take to reach the ground? (Round your answer to two decimal places.) s

Explanation / Answer

We have a stone in free fall, therefore gravity is the only force that affects the stone and therefore, g is the only acceleration that suffers the stone in the vertical direction.
a) With all this information at hand, we have that the position h(t) as a function of time is given by the following expression
h(t) = 550 - 4.9*t^2.
b) The time at which the stone hits the ground is found by making the height h(t) zero. This gives
550 = 4.9*t^2,
t = 10.59 s.
c) The velocity v(t) is found by derivating the position h(t) with respect to time. We obtain that
v(t) = h'(t) = - 9.8*t. Therefore, the velocity with which the stone reaches the ground is
v(10.59) = -9.8*10.59 = -103.78 m/s.
It is negative because it is directed downwards.
d) If there is an initial velocity, the equation of motion changes as
h(t) = 550 - 9*t - 4.9*t^2,
then the time that take it to reach the ground is obtained by making h=0 and solving for t.
This gives
t = 9.71 s     and    t=-11.55 s.
We take the first solution since it is the physically correct one.

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