This is a homework problem I was given. I know that somehow you have to use the

Question

This is a homework problem I was given. I know that somehow you have to use the physics motion equations with distance to find two equations for the x and y components and then solve the system, however I do not now how to go about this. The question is as follows:

A ball is kicked at an angle theta = 45degree. 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 d0 = 5 m. and the truck moves directly away from the ball at velocity V- 9 m/s (as shown), what is the maximum and minimum velocity of the ball, v0 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.

Explanation / Answer

Voy = original velocity of y

Vox = original velocity of x

Time that the ball will land is t = Voy / 0.5g = v sin45 / 4.9

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

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

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

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

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)

0 = 5 + 9([v(0.7071)]/4.9) - v(0.7071) ([v(0.7071)]/4.9)

divide 0.7071 by 4.9 in both the first and second part of the equation

0 = 5 + 9(v[.14431]) - v(0.7071) (v[.14431])

0 = 5 + 9v(.14431) - v2(.10204)

0 = 5 + 1.29879v - .10204 v2

ax2 + bx + c = 0

-0.10204 v2 + 1.2988 v + 5 = 0

v = 15.82 m/s [ minimum velocity]

For the maximum velocity, add 2.5 m to 5 m = 7.5 m

v = 17.04 m/s [maximum velocity]

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