Answer:
The equation of the position at time t will be:
[tex]y(t)=-\frac{1}{2}9.81t^{2}+64t+15[/tex]
Step-by-step explanation:
We can start by saying that the acceleration here is g = -9.81 m/s². The minus sign is because the gravity acceleration is a vector downward (negative value), and the ball is going upward (positive value).
And we know that acceleration can be written as a second derivative:
[tex]g=\frac{d^{2}y}{dt^{2}}=y''[/tex] (1)
Now, we can take the integral in each side:
[tex]\int gdt=\int y''dt[/tex]
[tex]\int y''dt=\int -9.81dt[/tex] (2)
Solving the integral in each side we have:
[tex]y'(t)=-9.81t+C[/tex] (3)
Where y'(t) is the velocity at t time (v = dy/dt = y' ) and c is a constant value.
Now, the initial conditions are:
initial height y(0) = 15 feet
initial velocity y'(0) = v(0) = 64 feet/s
Using this condition we can find C. Let's evaluate equation (3) at t = 0.
[tex]y'(0)=-9.81(0)+C[/tex]
[tex]C=64\: feet/s[/tex]
So we have:
[tex]y'(t)=-9.81t+64[/tex] (4)
Now, we need to take the integral of equation (4) to get the position function:
[tex]\int y'(t)dt=\int (-9.81t+64)dt[/tex]
Solving this new integral we have:
[tex]y(t)=-\frac{1}{2}9.81t^{2}+64t+D[/tex] (5)
Using the same method above, we can find D evaluating (5) at t = 0, we have:
[tex]y(0)=-\frac{1}{2}9.81(0)^{2}+64(0)+D[/tex]
[tex]D=y(0)[/tex]
[tex]D=15\: feet[/tex]
Finally, the equation of the position at time t will be:
[tex]y(t)=-\frac{1}{2}9.81t^{2}+64t+15[/tex]
I hope it helps you!
