Answer:
Horizontal
component of the velocity of the ball
\[u\,\,\cos 60{}^\circ =36\times
\frac{1}{2}=18\text{ }km/h\]
Vertical component of the
velocity of the ball
\[=u\text{ }\sin 60{}^\circ
=36\times \frac{\sqrt{3}}{2}=\text{ }31.18\text{ }km/h\]
Since the speed of
car is equal to the horizontal component of the velocity of the ball, so for
the boy sitting in the moving car, ball appears to move up vertically upward
and then move down vertically downward.
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