# physics

A golfer hits a ball off a tee from an elevated platform 13m above the ground. if the golf ball leaves the tee at 85 m/s @ 15 degrees, find the flight time, the maximum height above the ground and the horizontal displacement of the golf ball.

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1. Vo = 85m/s[15o]
Xo = 85*cos15 = 82.1 m/s.
Yo = 85*sin15 = 22 m/s.

Y = Yo + g8*Tr = 0 @ max ht.
22 - 9.8Tr = 0
9.8Tr = 22
Tr = 2.24 s. = Rise time.

0.5g*Tf^2 = h max = 37.7 m.
4.9Tf^2 = 37.7
Tf^2 = 7.69
Tf = 2.77 s. = Fall time.

T = Tr + Tf = 2.24 + 2.77 = 5 s. =
Flight time.

h max = ho + (Y^2-Yo^2)/2g
= 13 + (0-22^2)/-19.6 = 37.7 m.

Dx = Xo*T = 82.1m/s * 5s = 412 m.

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posted by Henry

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