Projectile from a moving vehicle

[dmkeddy]

Homework Statement

Joe is in a car traveling 90km/h, following Mark in a car 20m ahead traveling 110km/h. Joe then decides to throw a water balloon at Mark at an angle of 45 degrees in order to hit Mark. What speed should the water balloon be launched?

Vertical:
Vi = 14m/s
Vf = -14m/s
a = -9.8m/s^2
t =
d = 20m

Horizontal:
V = 14m/s
t =
d = 20m


I used the equation d(horizontal) = v^2/g x Sin2(theta) in order to find the horizontal vector component of 14m/s which is equal to the vertical vector component, as the degree is 45. This would mean the hypotenuse is 19.8 m/s, which would be the launching speed, but how would the velocity of the two cars affect it? As well as the further ahead car going a faster speed.

 

[jbsteele]

Homework Equations d(horizontal) = v^2/g x Sin2(theta)The Attempt at a SolutionThe launching speed should be 19.8m/s, regardless of the velocity of the two cars involved. The vertical vector component is equal to the horizontal vector component due to the angle of 45 degrees, so the speed of 19.8m/s is the same regardless of the speeds of the two cars.