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(B) Let the height of the tower be $h$ and the acceleration due to gravity be $g = 10 \ m/s^2$.For ball $A$ thrown upwards: $-h = ut_A - \frac{1}{2}gt

Vedclass · Accepted Answer

$60$

Vedclass · Answer

(B) Let the height of the tower be $h$ and the acceleration due to gravity be $g = 10 \ m/s^2$.For ball $A$ thrown upwards: $-h = ut_A - \frac{1}{2}gt_A^2 \implies -h = 6u - \frac{1}{2}(10)(6)^2 \implies -h = 6u - 180 \implies h = 180 - 6u$  (Equation $1$)For ball $B$ thrown downwards: $h = ut_B + \frac{1}{2}gt_B^2 \implies h = 2u + \frac{1}{2}(10)(2)^2 \implies h = 2u + 20$ (Equation $2$)Equating the two expressions for $h$: $180 - 6u = 2u + 20$$160 = 8u \implies u = 20 \ m/s$Substituting $u$ in Equation $2$: $h = 2(20) + 20 = 40 + 20 = 60 \ m$.Thus,the height of the tower is $60 \ m$.

Column $I$	Column $II$
$(A)$ Height of particle from ground	$(p)$ $0$
$(B)$ Speed of particle	$(q)$ $10$ $SI$ unit
$(C)$ Displacement of particle	$(r)$ $40$ $SI$ unit
$(D)$ Acceleration of particle	$(s)$ $20$ $SI$ unit

Two balls $A$ and $B$ are thrown with the same velocity $u$ from the top of a tower. Ball $A$ is thrown vertically upwards and ball $B$ is thrown vertically downwards. If $t_A = 6 \ s$ and $t_B = 2 \ s$,then the height of the tower is ......... $m$.

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