A hang glider pilot is flying horizontally over an open field when her shoe falls off. She was going 23.0 m/s and was 81.0 m above the ground. How long did it take for the shoe to hit the ground?
height=1/2 g*t^2
t=sqrt (2*81/9.8) s
To find the time it takes for the shoe to hit the ground, we can use the equations of motion. The shoe is falling vertically, so we only need to consider its vertical motion.
The equation that relates the initial vertical velocity (Vi), time (t), and displacement (h) is:
h = Vi * t + (1/2) * g * t^2
Where g is the acceleration due to gravity, which is approximately 9.8 m/s^2.
In this case, the initial vertical velocity of the shoe is 0 m/s because it was dropped vertically and not given any initial upward or downward velocity.
So, the equation simplifies to:
h = (1/2) * g * t^2
Rewriting the equation to solve for time:
t^2 = 2h / g
t = sqrt(2h / g)
Now we can substitute the values into the equation:
h = 81.0 m
g = 9.8 m/s^2
t = sqrt(2 * 81.0 / 9.8)
Calculating the result:
t = sqrt(16.5306)
t ≈ 4.07 seconds
Therefore, it takes approximately 4.07 seconds for the shoe to hit the ground.