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A long jumper leaves the ground at 45 ∘ above the horizontal and lands 9.1 m away. What is her takeoff speed v0?

Answer :

Answer:

So takeoff speed will be 13.35 m/sec    

Explanation:

We have given that horizontal range R = 9.1 m

Angle from horizontal at which jumper leaves the ground [tex]\Theta =45^{\circ}[/tex]

Acceleration due to gravity [tex]g=9.8m/sec^2[/tex]

We know that horizontal range is given by [tex]R=\frac{u^2sin2\Theta }{2g}[/tex]

[tex]9.1=\frac{u^2sin2\times 45 }{2\times 9.8}[/tex]

u = 13.35 m/sec

So takeoff speed will be 13.35 m/sec

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