Question

A ball is thrown upward from the top of a 200 foot tall building with a velocity of 40 feet per second. Take the positive direction upward and the origin of the coordinate system at ground level. What is the initial value problem for the the position, LaTeX: x\left(t\right)\:x ( t ), of the ball at time t

Answer 1

Initial Value Problem:
`(d^2 x)/(dt^2) = -32, x(0) = 200, (dx)/(dt)(0) = 40`

`x(t) = -16t^2 + 40t +200`

Explanation:

The ball is thrown vertically downward, this means that acceleration due to gravity,
`g = (dx^(2) )/(dt^(2) ) = - 32 ft/s^2`

The height of the ball at time, t = 0 at the top of the building will be:
`x(0) = 200 ft`

The velocity at which the ball is thrown from the top of the building,
`(dx)/(dt) (0)= 40 ft/s`

Therefore the initial value problem is written below:

`(d^2 x)/(dt^2) = -32, x(0) = 200, (dx)/(dt)(0) = 40`

Let us solve for x(t)

`(d^2 x)/(dt^2) = -32\\d((dx)/(dt) )= -32 dt\\`

Integrate both sides

`(dx)/(dt) = -32t + k_1\\(dx)/(dt) (0) = 40\\40 = -32(0) + k_1\\k_1 = 0\\(dx)/(dt) = -32t + 40`

Integrate both sides

`x(t) = -16t^2 + 40t + k_2\\x(0) = 200\\200 = -16(0) + 40(0) + k_2\\k_2 = 200\\x(t) = -16t^2 + 40t +200`