JEE MAIN - Physics (2022 - 27th July Morning Shift - No. 2)
A bullet is shot vertically downwards with an initial velocity of $$100 \mathrm{~m} / \mathrm{s}$$ from a certain height. Within 10 s, the bullet reaches the ground and instantaneously comes to rest due to the perfectly inelastic collision. The velocity-time curve for total time $$\mathrm{t}=20 \mathrm{~s}$$ will be:
(Take g = 10 m/s2).
_27th_July_Morning_Shift_en_2_1.png)
_27th_July_Morning_Shift_en_2_2.png)
_27th_July_Morning_Shift_en_2_3.png)
_27th_July_Morning_Shift_en_2_4.png)
Explanation
$$|{v_{10}}| = (100 + 10 \times 10)$$ m/s
$${v_{10}} = - 200$$ m/s and $${v_0} = - 100$$ m/s
from 10s to 20s velocity remains zero
$$\Rightarrow$$ from t = 0s to 10s velocity increases in magnitude linearly.
$$\Rightarrow$$ graph given in option A fits correctly.
Comments (0)
