Question

A ball is dropped from rest at point O . It passes a window with height 3.8 m in time interval tAB = 0.02 s.Identify the correct pair of equations, which enable us to solve for speed vB . Assume down is the positive y direction. Use g = 9.8 m/s 2 . k~vk = v is the speed of the ball.

Answer 1

VB − VA = g tAB & (VA + VB)/2 = h / tAB

Step-by-step explanation:

s = h = Displacement

tAB = t = Time taken

VA = u = Initial velocity

VB = v = Final velocity

a = g = Acceleration due to gravity = 9.8 m/s²

`s=ut+(1)/(2)at^2\\\Rightarrow u=(s-(1)/(2)at^2)/(t)\\\Rightarrow u=(3.8-(1)/(2)* 9.8* 0.02^2)/(0.02)\\\Rightarrow u=189.902\ m/s`

`v=u+at\\\Rightarrow v=189.902+9.8* 0.02\\\Rightarrow v=190.098\ m/s`

`(v+u)/(2)=(190.098+189.902)/(2)=190\ s`

`(h)/(t)=(3.8)/(0.02)=190\ s`

Hence, the equations VB − VA = g tAB & (VA + VB)/2 = h / tAB will be used