Question

2 bodies a and b of equal mass are projected simultaneously with speed 10m/s and with angle 30 find coordinate of center of mass m at t=10

Answer 1

Increase in vertical distance will be provided by mass projected vertically upwards .

Hence,

At Max height, v_y = 0

u_y = 20m/s \\ a_y = -10 m/s^2 \\ v_y^2 -u_y^2 = 2a_yS_y \\ => 0 - 20^2 = 2 * -10 * S_y \\ => S_y = 20 m

2) From Ground,

Maximum height achieved by centre of mass,

= 20+20 = 40m

3.6

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