Question

All the particles thrown with the same initial speed would strike the ground:

• A. With the same speed
• B. Simultaneously
• C. Time would be least for the particle thrown with velocity $v$ downwards, i.e., particle (1)
• D. Time would be maximum for the particle (2).

Answer 1

Answer: (A), (C), (D)

Time would be maximum for the particle (2).

Mechanical energy will be conserved in all cases.
In all situations K.E. (or, speed) at the ground are equal.
i.e., option a is correct.
$h=v{t}_1+\frac{1}{2}g{t}_1^2$ [for (first) particle] (i)
$h=v{t}_2-\frac{1}{2}g{t}_2^2$ [for (second) particle] (ii)
From equation (i) and equation (ii) $t_2>t_1\left(t_2\right)$ = maximum , $\left(t_1\right)$ = minimum
i.e., options (c) and (d) are correct.