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

Answer:

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

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

$s=ut+\frac{1}{2}at^2\\\Rightarrow u=\frac{s-\frac{1}{2}at^2}{t}\\\Rightarrow u=\frac{3.8-\frac{1}{2}\times 9.8\times 0.02^2}{0.02}\\\Rightarrow u=189.902\ m/s$

$v=u+at\\\Rightarrow v=189.902+9.8\times 0.02\\\Rightarrow v=190.098\ m/s$

$\frac{v+u}{2}=\frac{190.098+189.902}{2}=190\ s$

$\frac{h}{t}=\frac{3.8}{0.02}=190\ s$

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