Respuesta :
(a) [tex]y(t)=250 - 4.9 t^2[/tex]
For an object in free-fall, the vertical position at time t is given by:
[tex]y(t) = h + ut - \frac{1}{2}gt^2[/tex]
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
[tex]y(t) = 250 - \frac{1}{2}(9.8) t^2 = 250 - 4.9 t^2[/tex]
(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
[tex]y(t) = h - \frac{1}{2}gt^2=0[/tex]
So for the stone in the problem, we have
[tex]250 - 4.9 t^2 = 0[/tex]
Solving for t, we find:
[tex]t=\sqrt{\frac{250}{4.9}}=7.14 s[/tex]
(c) -70.0 m/s (downward)
The velocity of an object in free fall is given by the equation
[tex]v(t) = u - gt[/tex]
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:
[tex]v=0-(9.8 m/s^2)(7.14 s)=-70.0 m/s[/tex]
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
[tex]y(t) = h + ut - \frac{1}{2}gt^2=250 - 2t - 4.9 t^2[/tex]
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