- Home
- Standard 11
- Physics
3-2.Motion in Plane
hard
A person is standing on an open car moving with a constant velocity of $30\,\,m/s$ on a straight horizontal road. The man throws a ball in the vertically upward direction and it returns to the person after the car has moved $240\,\,m.$ The speed and the angle of projection
A
as seen from the car is $40\,\,m/s,\,\, 90^o$
B
as seen from the road is $50\,\,m/s,\,\,tan^{-1}\,\,(4/3)$
C
both $(A)$ and $(B)$
D
none
Solution
From ground
$\mathrm{R} =\frac{2 \mathrm{u} \times \mathrm{u}_{\mathrm{y}}}{\mathrm{g}} \Rightarrow 240=\frac{2 \times 30 \times \mathrm{u}_{\mathrm{y}}}{10} \Rightarrow \mathrm{u}_{\mathrm{y}}=40 \mathrm{m} / \mathrm{s}$
$\tan \theta=\frac{\mathrm{u}_{\mathrm{y}}}{\mathrm{u}_{\mathrm{x}}}=\frac{40}{30}=\frac{4}{3} $
Standard 11
Physics
Similar Questions
medium