Question

A ball is dropped from a building of height 45m .Simultaneously another ball is thrown up with a speed 40 m/s. Calculate the relative velocity of the balls as a function of time.​

Answer 1

v^2 = u^2 + 2gh

=0 +2×10×45

(g=10; initial velocity u=0)

v^2=20×45

v^2 =900

v = 30 m/s ....(taking sqareroot )

speed of ball dropped from building is 30 m/s and the ball is thrown upward with velocity 40 m/s.

relative velocity of 1st ball,

v12 = v1 - v2

= 30-40

= -10 m/s

reltive velocity of a ball 1 wrt ball 2 is -10 m/s

relative velocity of a 2nd ball

v21 = v2 - v1

= 40 - 30

= 10 m/s

relative velocity of a ball 2 wrt ball 1 is 10 m/s

Answer 2

Answer:

70m/s but as a function of time it is t^0

Explanation:

Given in the attachment

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