Tennis Ball Cannon -- height of ascent?

I'll post a solution tonight, but it should be ~31 m.

That is assuming Drag ∝ v^2 at all points.

(Need to check my math when I have time).

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I. Find the equation for the forces acting on the ball:

F = ma

a = F/m

F = Fg + Fd

Fg = mg

Fd = F drag

Fd = Cv^2 ... I am making the assumption that drag varies solely with v^2. This is a fairly good assumption because it covers the regions of highest drag

C = Fd/v^2

But, we know that at terminal velocity (v_t) the drag force = gravity, i.e.

Fd = Fg

mg = C v_t^2

C = mg/v_t^2, ergo

m a = Fg + Fd

m a = mg + mg(v/v_t)^2

II. Find the equation of motion in terms of velocity and time.

m a = mg + mg(v/v_t)^2

a = g(1 + (v/v_t)^2)

dv/dt =g(v_t^2 + v^2)/ v_t^2)

v_t^2/(v_t^2 + v^2) dv = g dt

atan (v/v_t) = g/v_ t + A ... A is a constant

v = v_t * tan (g/v_t + A)

but we know that v_o = 2 v_t

v_o = v_t * tan (g*(0)/v_t + A)

2v_t = v_t tan (A)

2 = tan (A)

A = atan (2)

so our equation of motion based on v is:

v = v_t tan (atan(2) - g t/v_t )

[Note: I flipped the sign to make everything kosher in the frame of reference]

>>>Before we go on, we need to find the time to reach max height -- this is when the velocity = 0

v= 0 = v_t tan (atan(2) - g t/v_t )

atan(2) - g t/v_t = 0

t = v_t/g atan(2)

Here v_t = 20 m/s and g = 10 m/s^2, ergo

t = 2 atan(2) = 2.2143 s <<<<<<

III. Find the equation of motion in terms of x and t:

Retrieving our equation of motion from above:

v = v_t tan (atan(2) - g t/v_t)

dx/dt = v_t tan (atan(2) - g t/v_t)

dx = v_t tan (atan(2) - g t /v_t) dt

x = v_t ln(cos(atan(2) - g t/v_t) / g/v_t + B ... B is a constant

x = v_t^2/g ln(cos(atan(2) - g t/v_t) + B

But we know that

x = f(0) = 0

B = - v_t^2/g ln(cos(atan(2)), ergo

x = v_t^2/g (ln(cos(atan(2) - g t/v_t) - ln (cos (atan(2))

IV. Plug n' chug:

From above, at max, t = v_t/g atan(2)

x.max = v_t^2/g (-ln(cos(atan(2)))

x.max = 32.1898 m

H = Vi * t -(1/2) g t^2 ***(-) since it is going up

Note that if we graph H for vertical values and t for horizontal values

the tangent at any given time = H/t which defines velocity at any given time in the motion

so if we take the first derivative of our equation we get

dH/dt = Vi - 2(1/2) g t ****this is our formula for velocity at any given time

Note that by the time gravity consumes the momentum of the initial velocity the ball start to free fall. That means it has reach its maximum where the slope is 0 and so is our velocity.

Therefore if we plug in a value of 0 for our velocity, we can then compute the time when the ball stops rising and before it starts falling

0 = Vi - g t^2

0 = 40 - 10 t^2

t = (40/10)

t = 4

Plug it back into the H equation

H = (40)(4) - (1/2)(10)(4^2)

H = 160 - 80

H = 80 m *****ANSWER

final velocity = initial velocity + acceleration * time

at the top, final velocity = 0

0 = 40 + (-10)*t

t = 40 / 10 ~ 4 seconds

distance = (initial velocity)*time + (1/2)*acceleration*(time^2)

height = 40*4 + 0.5*(-10)*16)

height = 160 - 80,

height = 80 meters

Haha I am sorry but I have to agree with the guy above me.