Vector Analysis of Throwing a Ball from a Moving Car to a Stationary Target

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To throw the ball straight east to the girl while in a car moving north at 10 km/h, the boy must throw the ball at an angle that compensates for the car's velocity. The ball's velocity relative to the car is 20 km/h, and the northward motion of the car adds a component to the ball's trajectory. By using vector components, a right triangle can be established to determine the required angle and the resultant velocity of the ball relative to the ground. The total velocity of the ball can be calculated using the Pythagorean theorem, combining the eastward throw and the northward movement. Understanding these vector components is essential for accurately determining the ball's path and speed.
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Vector Question please help

Homework Statement


A boy in a car moving at 10km/h[N] wants to throw a ball to a girl standing on the right-hand side of the road. He is able to throw the ball at 20km/h. He wants the ball to go straight east, directly to her.
(a) Relative to the car, which way should he throw the ball in order to do this?
(b) Relative to the ground, how fast will the ball travel to reach her?

Homework Equations


not sure (missed class)

The Attempt at a Solution


not sure (missed class)
 
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Im not comletely positive if this is right but, I think that they are giving you components of the balls velocity. The ball is moving 10 km/h N due to being in the car, and the boy can throw the ball 20 km/h E. From their you can set up a right triangle to find the angle that the ball will be moving relative to the car, and you can also find the hypotenuse which would be the balls total velocity. Again I am not completely sure, but i hope this helps.
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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