A ball is dropped from the top of a building | vedclass

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(A) Let the height of the building be $H = 80 \,m$.For the ball dropped from the top: Initial velocity $u_1 = 0$,acceleration $a_1 = -g$.For the ball

Vedclass · Accepted Answer

$1.6$

Vedclass · Answer

(A) Let the height of the building be $H = 80 \,m$.For the ball dropped from the top: Initial velocity $u_1 = 0$,acceleration $a_1 = -g$.For the ball thrown from the bottom: Initial velocity $u_2 = 50 \,m/s$,acceleration $a_2 = -g$.The relative acceleration between the two balls is $a_{	ext{rel}} = a_2 - a_1 = (-g) - (-g) = 0$.The relative velocity between the two balls is $v_{	ext{rel}} = u_2 - u_1 = 50 - 0 = 50 \,m/s$.Since the relative acceleration is zero,the time taken for them to meet is given by $t = \frac{H}{v_{	ext{rel}}}$.Substituting the values: $t = \frac{80}{50} = 1.6 \,s$.

$A$ ball is dropped from the top of a building of height $80 \,m$. At the same instant,another ball is thrown upwards with a speed of $50 \,m/s$ from the bottom of the building. The time at which the balls will meet is ........ $s$.

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