Astone is dropped from the upper observation deck of a tower, 250 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 2 m/s, how long does it take to reach the ground? (round your answer to two decimal places.) s

(a)

For an object in free-fall, the vertical position at time t is given by:

where

h is the initial vertical position

u is the initial vertical velocity

g = 9.8 m/s^2 is the acceleration of gravity

t is the time

In this problem,

h = 250 m

u = 0 (the stone starts from rest)

So, the vertical position of the stone is given by

(b) 7.14 s

The time it takes for the stone to reach the ground is the time t at which the vertical position of the stone becomes zero:

y(t) = 0

Which means

So for the stone in the problem, we have

Solving for t, we find:

(c) -70.0 m/s (downward)

The velocity of an object in free fall is given by the equation

where

u is the initial velocity

g = 9.8 m/s^2 is the acceleration of gravity

t is the time

Here we have

u = 0

So if we substitute t = 7.14 s, we find the velocity of the stone at the time it reaches the ground:

The negative sign means the direction of the velocity is downward.

(d) 6.94 s

In this situation, the stone is thrown downward with an initial speed of 2 m/s, so its initial velocity is

u = -2 m/s

So the equation of the vertical position of the stone in this case is

By solving the equation, we find the time t at which the stone reaches the ground.

We find two solutions:

t = -7.35 s

t = 6.94 s

The first solution is negative, so it has no physical meaning, therefore we discard it. So, the time it takes for the stone to reach the ground is:

t = 6.94 s