Question

A hammer falls from a construction worker's hand 30m above the ground at the same time that a filled soft drink can is projected vertically upward from the ground at a velocity of 24 m/s. Where and when will the two objects meet?

Answer 1

For harmer:-

initial velocity=u=0m/s

Acceleration due to gravity=10m/s^2

Now

$\\ \sf\longmapsto s=ut+\dfrac{1}{2}gt^2$

$\\ \sf\longmapsto s=0t+\dfrac{1}{2}(10)t^2$

$\\ \sf\longmapsto s=5t^2\dots(1)$

For soft drink:-

u=24m/s

$\\ \sf\longmapsto 30-s=(24)t+\dfrac{1}{2}(10)t^2$

Using eq(1)

$\\ \sf\longmapsto 30-5t^2=24t-5t^2$

$\\ \sf\longmapsto 24t=30$

$\\ \sf\longmapsto t=\dfrac{30}{24}$

$\\ \sf\longmapsto t=1.2s$

After 1.2s they will meet.

Now

putting t in eq(1)

$\\ \sf\longmapsto s=5t^2=5(1.2)^2=5(1.44)=7.2m$

At the height of 7.2m they will meet