Asked by Alex
On a flat golf course, a golf ball lies 12m from the base of a lamp post. A golfer hits the ball with a club, giving the golf ball an initial velocity of 18 m/s. At 2.9 seconds, after being hit, the golf ball strikes the top of the lamp post. How high (m) is the lamp post?
Answers
Answered by
Damon
It goes up a distance h in 2.9 seconds
It goes horizontal a distance 12 meters in 2.9 seconds
so
u = 18 cos T = 12/2.9
cos T = .23
T = 76.7 degrees above horizontal
so initial velocity up Vi = 18 sin 76.7
= 17.5 m/s
h = Vi t -.5*9.81*t^2
h =17.5*2.9 - 4.9(2.9)^2
= 50.8-41.2
= 9.59 meters
It goes horizontal a distance 12 meters in 2.9 seconds
so
u = 18 cos T = 12/2.9
cos T = .23
T = 76.7 degrees above horizontal
so initial velocity up Vi = 18 sin 76.7
= 17.5 m/s
h = Vi t -.5*9.81*t^2
h =17.5*2.9 - 4.9(2.9)^2
= 50.8-41.2
= 9.59 meters
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