Deriving Equations - Freefall, Velocity

[dragon_dance]

Homework Statement

Given a Height of release (d) and the Total time (T) form hand to floor. Derive general equations for vi, vf, h (in terms of d, T, and g)

vi = velocity ball leaves hand
vf = velocity of the ball the instant before it strikes the ground
h = height of the ceilingknown
height of ceiling = 9.67 meters [including height of release]
height of release = 1.58 meters
velocity ball leaves hand = 12.59 m/s
velocity ball hits floor = 13.77 m/s
time it would take to fall from ceiling to floor = 1.41 seconds

Homework Equations

The Attempt at a Solution

Just confused at how to start deriving this..

 

[Jhero]

To derive the equations for vi, vf, and h, we can use the equations of motion for constant acceleration. These equations are:

vf = vi + at
h = vi*t + (1/2)*a*t^2
vf^2 = vi^2 + 2ah

We can also use the equation for acceleration due to gravity, which is g = 9.8 m/s^2.

To start, we can use the given information to solve for the acceleration (a) of the ball as it falls from the hand to the floor. We can use the equation for h to solve for a:

a = (2*(h - d))/T^2

Substituting in the known values for h, d, and T, we get:

a = (2*(9.67 - 1.58))/(1.41)^2
a = 8.61 m/s^2

Now, we can use this value for a to solve for vi, vf, and h.

For vi:
Using the equation vf = vi + at, we can solve for vi:

vi = vf - at
vi = 13.77 - (8.61)*(1.41)
vi = 1.81 m/s

For vf:
Using the equation vf^2 = vi^2 + 2ah, we can solve for vf:

vf = √(vi^2 + 2ah)
vf = √(1.81^2 + 2*(8.61)*(9.67 - 1.58))
vf = 13.77 m/s

For h:
Using the equation h = vi*t + (1/2)*a*t^2, we can solve for h:

h = vi*t + (1/2)*a*t^2
h = (1.81)*(1.41) + (1/2)*(8.61)*(1.41)^2
h = 9.67 m

Therefore, the general equations for vi, vf, and h in terms of d, T, and g are:

vi = vf - (g*T)
vf = √(vi^2 + 2*g*(h - d))
h = (vi*T) + (1/2)*g*T^2