Question #108411
A 50kg track athlete running at 10.0 m/s jumps over a hurdle. If athlete’s horizontal component of the velocity over the hurdle is 8.8 m/s and the air resistance is disregarded, how high was the jump?
1
Expert's answer
2020-04-08T10:30:00-0400

As per the given question,

Mass of the athlete (m)=50 kg

Speed of the athlete (v)=10 m/sec

Let the angle of projection =θ=\theta

Let the jump was h meter high.

Horizontal component of the velocity =vcosθ=8.8m/sec=v \cos\theta=8.8 m/sec

10cosθ=8.8\Rightarrow 10 \cos\theta =8.8

cosθ=8.810=0.88\Rightarrow \cos \theta =\dfrac{8.8}{10}=0.88

θ=28.35\Rightarrow \theta =28.35^\circ

Vertical component of the velocity=vsinθ=10sin28.35=v\sin\theta=10 \sin 28.35^\circ

h=v2sin2θ2g\Rightarrow h=\dfrac{v^2\sin^2\theta}{2g}

h=10×10sin228.359.8\Rightarrow h=\dfrac{10\times 10 \sin^2 28.35^\circ}{9.8}

h=2.3m\Rightarrow h=2.3 m


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