The following three questions pertain to the following situation. A man is throw

Question

The following three questions pertain to the following situation.

A man is throwing a ball in the air as he is walking on the sidewalk. The man is walking in the positve x-direction with speed |v mg|= 2 m/s with respect to the ground. The initial velocity, v bm, of the ball with respect to the man is pointing in the positive y-direction with magnitude 7.5 m/s (see figure.) The velocity of the ball with respect to the ground, v bg, is given by v bg=v bm+v mg

What is the magnitude, |v bg| of the initial velocity of the ball with respect to the ground?

What is angle, , between v bg and the positive x-direction?

A woman is sitting at a cafe on the opposite side of the street, observing the man throwing the ball. What is the speed of the ball as seen from the woman?

Explanation / Answer

Speed of man wrt ground = Vmg = 2m/s
Velocity of Ball wrt Man in Y direction Vbm = 7.5m/s

Velocity of Ball wrt Ground Vbg = Vmg + Vbm
Velocity of Ball wrt Ground Vbg = 2 i + 7.5 j m/s

Magnitude of initial Velocity of Ball wrt Ground = sqrt(2^2 + 7.5^2) m/s
Magnitude of initial Velocity of Ball wrt Ground |Vbg| = 7.76 m/s

Direction = tan^-1(7.5/2)
= 75.070 anticlockwise from +ve X direction

Speed of the ball as seen from the woman = Magnitude of Initial Velocit of Ball wrt Ground = 7.6 m/s

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