Two balls of equal masses are projected upwards simultaneously, one from the ground with initial velocity $50 {ms}^{- 1}$ , and the other from a $40 m$ tower with initial velocity of $30 {ms}^{- 1} .$ The maximum height attained by their COM will be:
( $g = 10 {ms}^{- 2}$ )

Question

Two balls of equal masses are projected upwards simultaneously, one from the ground with initial velocity

50 {ms}^{- 1}

, and the other from a

40 m

tower with initial velocity of

30 {ms}^{- 1} .

The maximum height attained by their COM will be:
(

g = 10 {ms}^{- 2}

)

Toppr Admin · Accepted Answer

VCM-m1v1-m2v2m1-m2-v1-v2-2Initially-UCM-40m-sXCM-40-2-20maboveground-MaximumheightmeansVCM-0Thatis-02-x2212-402-2-xD7-x2212-10-xD7-H-x2212-20-ThusH-100m

Two balls of equal masses are projected upward simultaneously, one from the ground with speed $50 m / s$ and other from a $40 m$ high tower with initial speed $30 m / s$ . Find the maximum height attained by their centre of mass.
100 m
150 m
200 m
50 m

$V_{C M} = \frac{m_{1} v_{1} + m_{2} v_{2}}{m_{1} + m_{2}} = (v_{1} + v_{2}) / 2$
$I n i t i a l l y, U_{C M} = 40 m / s X_{C M} = 40 / 2 = 20 m a b o v e g r o u n d . M a x i m u m h e i g h t m e a n s V_{C M} = 0 T h a t i s, 0^{2} - 40^{2} = 2 \times (- 10) \times (H - 20) T h u s H = 100 m$

Two balls of equal masses are projected upward simultaneously, one from the ground with speed 50m/s and other from a 40m high tower with initial speed 30m/s. Find the maximum height attained by their centre of mass.100 m150 m200 m50 m

VCM=m1v1+m2v2m1+m2=(v1+v2)/2Initially,UCM=40m/sXCM=40/2=20maboveground.MaximumheightmeansVCM=0Thatis,02−402=2×(−10)×(H−20)ThusH=100m

Two balls of equal masses are projected upward simultaneously, one from the ground with speed $50 m / s$ and other from a $40 m$ high tower with initial speed $30 m / s$ . Find the maximum height attained by their centre of mass.
100 m
150 m
200 m
50 m

$V_{C M} = \frac{m_{1} v_{1} + m_{2} v_{2}}{m_{1} + m_{2}} = (v_{1} + v_{2}) / 2$
$I n i t i a l l y, U_{C M} = 40 m / s X_{C M} = 40 / 2 = 20 m a b o v e g r o u n d . M a x i m u m h e i g h t m e a n s V_{C M} = 0 T h a t i s, 0^{2} - 40^{2} = 2 \times (- 10) \times (H - 20) T h u s H = 100 m$