Question #135079
A football is kicked with an initial velocity of 29 m/s at an angle of 45 degrees with the horizontal. Determine the time of flight,horizontal distance,and the peak height of the football
1
Expert's answer
2020-09-28T08:09:15-0400

Given:

v0=29m/sv_0=29\:\rm m/s

θ=45\theta=45^{\circ}

(a) the time of flight

t=2v0sinθg=2×29×2/29.8=4.2st=\frac{2v_0\sin\theta}{g}=\frac{2\times29\times \sqrt{2}/2}{9.8}=4.2\:\rm s

(b) the horizontal distance

R=v02sin2θg=292×19.8=85.8mR=\frac{v_0^2\sin2\theta}{g}=\frac{29^2\times 1}{9.8}=85.8\:\rm m

(c) the peak height

H=v02sin2θ2g=292×1/22×9.8=21.5mH=\frac{v_0^2\sin^2\theta}{2g}=\frac{29^2\times 1/2}{2\times9.8}=21.5\:\rm m

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