williampalacek
Answered

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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

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