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?
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
To determine the height of the lamp post, we can use the equations of motion for vertical motion and the kinematic equation.
The kinematic equation for vertical motion is:
h = v0t + (1/2)gt^2
Where:
h is the height of the object (lamp post)
v0 is the initial vertical velocity (0 m/s because the ball only has horizontal velocity)
t is the time of flight (2.9 seconds)
g is the acceleration due to gravity (9.8 m/s^2)
The equation can be simplified as:
h = (1/2)gt^2
Substituting the given values:
h = (1/2)(9.8 m/s^2)(2.9 s)^2
Calculating:
h = (1/2)(9.8 m/s^2)(8.41 s^2)
h = 41.68 m
Therefore, the height of the lamp post is approximately 41.68 meters.