A) 436 ft
B) t = 5 sec
C) 5.60 s
Step-by-step explanation:
A)
The height of the ball at time t is given by the equation
[tex]h(t)=-16t^2+500[/tex]
where
[tex]-16 ft/s^2[/tex] is the acceleration of the ball (acceleration of gravity, downward)
+500 is the initial height of the ball, at time t = 0
Here we want to find the height of the ball after 2 seconds, so at a time of
t = 2 s
Substituting into the equation, we find:
[tex]h(2)=-16\cdot 2^2+500=436 ft[/tex]
B)
Here we want to find the time it takes for the ball to fall to a height of 100 feet above the ground, so the time t at which
h(t) = 100 ft
As stated in the text of the question, the height of the ball at time t is given by
[tex]h(t)=-16t^2+500[/tex]
Since
h(t) = 100, we have
[tex]100=-16t^2+500[/tex]
And solving for t we find:
[tex]16t^2=400\\t^2=25\\t=\pm 5[/tex]
So, the correct solution is the positive one:
t = 5 sec
C)
The ball reaches the ground when the height of the ball has became zero:
[tex]h(t) = 0[/tex]
The height of the ball at time t is given by
[tex]h(t)=-16t^2+500[/tex]
And substituting
h(t) = 0
We get
[tex]0=-16t^2+500[/tex]
And solving the equation for t, we find the time t at which the ball reaches the ground:
[tex]16t^2=500\\t^2=31.25\\t=\pm 5.6 s[/tex]
So, the correct solution is the positive one:
t = 5.60 s
