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Question

(A) Let the distance from the platform where the two balls meet be $x$.For the first ball:Initial velocity $u_1 = 0$,time $t_1 = 18\,s$,acceleration $

Vedclass · Accepted Answer

$75$

Vedclass · Answer

(A) Let the distance from the platform where the two balls meet be $x$.For the first ball:Initial velocity $u_1 = 0$,time $t_1 = 18\,s$,acceleration $g = 10\,m/s^2$.Using the equation of motion $h = ut + \frac{1}{2}gt^2$:$x = 0 	imes 18 + \frac{1}{2} 	imes 10 	imes (18)^2 = 5 	imes 324 = 1620\,m$.For the second ball:Initial velocity $u_2 = v$,time $t_2 = 18 - 6 = 12\,s$,acceleration $g = 10\,m/s^2$.Using the equation of motion $h = ut + \frac{1}{2}gt^2$:$x = v 	imes 12 + \frac{1}{2} 	imes 10 	imes (12)^2 = 12v + 5 	imes 144 = 12v + 720$.Equating the two expressions for $x$:$1620 = 12v + 720$$12v = 1620 - 720 = 900$$v = \frac{900}{12} = 75\,m/s$.

$A$ ball is dropped from a high-rise platform at $t=0$ starting from rest. After $6$ seconds,another ball is thrown downwards from the same platform with a speed $v$. The two balls meet at $t=18\,s$. What is the value of $v$ in $m/s$? (Take $g= 10\,m/s^2$)

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