Kicking A Ball Into The Back Of A Moving Truck - Solved

A ball is kicked at an angle θ = 45°. It is intended that the ball lands in the back of a moving truck which has a trunk of length L = 2.5 m. If the initial horizontal distance from the back of the truck to the ball, at the instant of the kick, is 5 m, and the truck moves directly away from the ball at velocity V = 9 m/s, what is the maximum and minimum velocity (v) so that the ball lands in the trunk. Assume that the initial height of the ball is equal to the height of the ball at the instant it begins to enter the trunk.

Voy = original velocity of y
Vox = original velocity of x
t = time

this drove me crazy but i finally figured it out.

time that the ball will land is t = Voy/.5g or:

v sin45/4.9

(4.9 is 1/2g [g = gravity at 9.8 m/s])

the distance that the ball will be (distance-x) for minimum velocity is Voxt = 5 + 9t or:

v cos45(v sin45/4.9) = 5 + 9(v sin45/4.9)

distance-x for maximum velocity is Voxt = 2.5 + 5 +9t or:

v cos45(v sin45/4.9) = 2.5 + 5 + 9(v sin45/4.9)

we'll start with the distance equation to solve for the minimum velocity first. turn the motherfucker into a quadratic and solve for v this way:

5 + 9(v sin45/4.9) = v cos45(v sin45/4.9)

0 = 5 + 9(v sin45/4.9) - v cos45(v sin45/4.9)

let's put in the numbers we know:

0 = 5 + 9([v(.7071)]/4.9) - v(.7071) ([v(.7071)]/4.9)

v cos45 and v sin45 are the same.

now we can divide .7071 by 4.9 in both the first and second part of the equation:

0 = 5 + 9(v[.14431]) - v(.7071) (v[.14431])

in the first part we then have 9v multiplied by .14431, and in the second part we can multiply the 2 v's together as well as .7071 and .14431 to get:

0 = 5 + 9v(.14431) - v^2(.10204)

then we multiply 9v by .14431, and v^2 by .10204:

0 = 5 + 1.29879v - .10204v^2

here's our fucking quadratic.

let's rearrange it so it looks like this familiar cocksucker:

ax^2 + bx + c = 0

so that we have:

-.10204v^2 + 1.2988v + 5 = 0

then we can use the following formula:

to solve for v.

plugging in the numbers we find:

v = 15.82 m/s for minimum velocity

for the maximum velocity, just add 2.5 m to 5 m = 7.5 m

so then

v = 17.04 m/s

THERE!

*note - for shits and giggles, try it with different starting angles and/or different starting positions and velocities for the truck. if you remember the equation for range time -  (v sin angle)/4.9 - you're in business.