williampalacek
Answered

Welcome to Westonci.ca, where curiosity meets expertise. Ask any question and receive fast, accurate answers from our knowledgeable community. Get immediate and reliable solutions to your questions from a community of experienced professionals on our platform. Join our Q&A platform to connect with experts dedicated to providing accurate answers to your questions in various fields.

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?

Sagot :

For harmer:-

  • initial velocity=u=0m/s
  • Acceleration due to gravity=10m/s^2

Now

[tex]\\ \sf\longmapsto s=ut+\dfrac{1}{2}gt^2[/tex]

[tex]\\ \sf\longmapsto s=0t+\dfrac{1}{2}(10)t^2[/tex]

[tex]\\ \sf\longmapsto s=5t^2\dots(1)[/tex]

For soft drink:-

  • u=24m/s

[tex]\\ \sf\longmapsto 30-s=(24)t+\dfrac{1}{2}(10)t^2[/tex]

  • Using eq(1)

[tex]\\ \sf\longmapsto 30-5t^2=24t-5t^2[/tex]

[tex]\\ \sf\longmapsto 24t=30[/tex]

[tex]\\ \sf\longmapsto t=\dfrac{30}{24}[/tex]

[tex]\\ \sf\longmapsto t=1.2s[/tex]

  • After 1.2s they will meet.

Now

putting t in eq(1)

[tex]\\ \sf\longmapsto s=5t^2=5(1.2)^2=5(1.44)=7.2m[/tex]

  • At the height of 7.2m they will meet

We hope you found what you were looking for. Feel free to revisit us for more answers and updated information. We appreciate your time. Please come back anytime for the latest information and answers to your questions. Get the answers you need at Westonci.ca. Stay informed with our latest expert advice.