Assume that the ball is thrown with a speed of v 2 = 8 m/s (with respect to the

Question

Assume that the ball is thrown with a speed of v2 = 8 m/s (with respect to the car) and an angle ?= 30o to the y-axis (as shown), also in the reference frame of the car. The car has a speed v1 = 12 m/s.

What is the speed of the ball with respect to the ground?

v = 19.3 m/s

v = 32.2 m/s

v = 14.4 m/s

v = 20 m/s

v = 17.4 m/s

If the speed that the ball is thrown with (with respect to the car) is increased, keeping all other parameters the same, how does the final speed of the ball with respect to the ground change?

vb,g increases.

vb,g would stay the same.

vb,g decreases.

Explanation / Answer

Vcg (velocity of car with respect to ground) = 12 x m/s
Vbc (velocity of ball with respect to car) = 8 sin 30 x + 8 cos 30 y
= 4 x + 6.93 y

Vbg (velocity of ball with respect to ground) =Vbc + Vcg
= 4 x + 6.93 y +12x
= 16 x + 6.93 y
magnitude = sqrt (16^2 + 6.93^2)
= 17.4 m/s

part B:
Increasing the speed will increase the final speed because speed of ball and speed ofcar adds up.
vb,g increases.

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