a) [tex]v(t)=8 -32.2t[/tex] [ft/s]
b) [tex]h(t)=185+8t-16.1t^2[/tex] [ft]
c) 186 ft at 0.248 s
Explanation:
a)
The motion of the weight is a free fall motion, which is subjected only to the force of gravity, downward.
Therefore, the motion of the object will be a uniformly accelerated motion, with constant acceleration of
[tex]g=32.2 ft/s^2[/tex] (acceleration due to gravity)
in the downward direction.
Therefore, the velocity of the weight after time t will be given by the following suvat equation:
[tex]v(t)=u-gt[/tex]
where:
[tex]u=+8 ft/s[/tex] is the initial vertical velocity (upward)
t is the time
Substituting the values of u and g, we find
[tex]v(t)=8 -32.2t[/tex] [ft/s]
b)
Since the motion of the weight is a uniformly accelerated motion, the expression that gives the displacement of the object at time t (and therefore, its height at time t), is given by
[tex]h(t)=h_0 + ut -\frac{1}{2}gt^2[/tex]
where
[tex]h_0 = 185 ft[/tex] is the initial height of the weight
[tex]u=+8 ft/s[/tex] is the initial vertical velocity (upward)
[tex]g=32.2 ft/s^2[/tex] (acceleration due to gravity)
t is the time
And substituting these values, we find:
[tex]h(t)=185+8t-16.1t^2[/tex] [ft]
c)
The time at which the weight reaches its maximum height above the ground can be found from the equation of the velocity: in fact, at the moment of maximum height, the direction of the weight changes, so its velocity will be zero at that instant:
[tex]v(t)=0[/tex]
So we have
[tex]0=u-gt[/tex]
And solving for t,
[tex]t=\frac{u}{g}=\frac{8}{32.2}=0.248 s[/tex]
So, the weight reaches its maximum height after 0.248 s.
In order to find the maximum height, we just need to substitute this time into the equation of the height, h(t).
By doing so, we find:
[tex]h(t)=h_0 + ut -\frac{1}{2}gt^2=185+8\cdot 0.248 -16.1\cdot (0.248)^2=186 ft[/tex]
