JEE MAIN - Physics (2022 - 30th June Morning Shift - No. 1)
At t = 0, truck, starting from rest, moves in the positive x-direction at uniform acceleration of 5 ms$$-$$2. At t = 20 s, a ball is released from the top of the truck. The ball strikes the ground in 1 s after the release. The velocity of the ball, when it strikes the ground, will be :
(Given g = 10 ms$$-$$2)
Explanation
_30th_June_Morning_Shift_en_1_1.png)
At t = 20 s,
velocity of truck,
v = 0 + 5 $$\times$$ 20 = 100 m/s
At 20 sec a ball is dropped from the truck, so velocity of ball will be same as truck.
Velocity of truck at x-direction = 100 m/s and in y-direction = 0.
$$\therefore$$ Velocity of ball vx = 100 m/s, vy = 0
Now ball will show projectile motion where vertically downward acceleration g = 10 m/s act on the ball.
As horizontally no acceleration acting on the ball so horizontal velocity 100 m/s will remain unchanged.
Velocity of the ball when it reach the ground along y-direction after 1 sec.
$${v_y} = 0 - 10 \times 1$$
$$\Rightarrow$$ $${v_y} = - 10$$ m/s
$$\therefore$$ Velocity of ball $$(\overrightarrow v ) = 100\widehat i - 10\widehat j$$
Comments (0)
