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+(1)/(2)gt^2`

`\\ \sf\longmapsto s=0t+(1)/(2)(10)t^2`

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

For soft drink:-

• u=24m/s

`\\ \sf\longmapsto 30-s=(24)t+(1)/(2)(10)t^2`

• Using eq(1)

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

`\\ \sf\longmapsto 24t=30`

`\\ \sf\longmapsto t=(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