Question

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A diver jumps up off the 10 meter tower and spends 1.63 seconds in the air. How fast was she going when her feet left the diving tower? How fast is she going just before she hits the water?

The diver has a speed of __________ as she leaves the tower.
The diver has a speed of __________ just before she hits the water.

Answer 1

• Height=h=10m
• Time=t=1.63s
• acceleration due to gravity=g=10m/s^2
• initial velocity=u
• Final velocity=v

Using 2nd equation of kinematics

`\\ \sf\longmapsto h=ut+(1)/(2)gt^2`

`\\ \sf\longmapsto 10=1.63t+(1)/(2)(10)(1.63)^2`

`\\ \sf\longmapsto 10=1.63u+5(2.65)`

`\\ \sf\longmapsto 10=1.63u+13.25`

`\\ \sf\longmapsto 1.63u=|-3.25|=3.25`

`\\ \sf\longmapsto u=2m/s`

Now

using 1st equation of kinematics

`\\ \sf\longmapsto v=u+gt`

`\\ \sf\longmapsto v=2+10(1.63)`

`\\ \sf\longmapsto v=2+16.3`

`\\ \sf\longmapsto v=18.3m/s`