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(A) Let the first ball be $B_1$ and the second ball be $B_2$.For $B_1$,initial velocity $u_1 = 5 \ m/s$. The total time taken to reach the collision p

Vedclass · Accepted Answer

$40$

Vedclass · Answer

(A) Let the first ball be $B_1$ and the second ball be $B_2$.For $B_1$,initial velocity $u_1 = 5 \ m/s$. The total time taken to reach the collision point is $t_1 = 2 \ s + 2 \ s = 4 \ s$.The distance covered by $B_1$ is $h = u_1 t_1 + \frac{1}{2} g t_1^2 = 5(4) + \frac{1}{2}(10)(4^2) = 20 + 80 = 100 \ m$.For $B_2$,initial velocity is $u_2$ and the time taken to reach the collision point is $t_2 = 2 \ s$.The distance covered by $B_2$ must be the same,so $h = u_2 t_2 + \frac{1}{2} g t_2^2$.Substituting the values: $100 = u_2(2) + \frac{1}{2}(10)(2^2)$.$100 = 2u_2 + 20$.$2u_2 = 80$.$u_2 = 40 \ m/s$.

$A$ ball is thrown vertically downward with a velocity of $5 \ m/s$. With what velocity should another ball be thrown downward after $2 \ s$ so that it can hit the $1^{st}$ ball in $2 \ s$?

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