Extreme-sports enthusiasts have been known to jump off the top of El Capitan, a sheer granite cliff of height 910m in Yosemite National Park. Assume a jumper runs horizontally off the top of El Capitan with speed 5.0m/s and enjoys a freefall until she is 150m above the valley floor, at which time she opens her parachute .
Q: how long is the jumper in freefall? ignore air resistance
x=1/2*g*t^2
x=910-150=760m
a=9,81m/s^2
t=sqrt(x/(0.5*g))
that is t=sqrt(760/(0.5*9,81)=12,44s=12s
sqrt(2x760)/9.8 = 12s
seweqeqe
To determine how long the jumper is in freefall, we can use the equations of motion for freefall. In this scenario, the jumper runs horizontally off the top of El Capitan with a speed of 5.0m/s, and then falls freely until she is 150m above the valley floor.
First, let's consider the vertical motion of the jumper. We know that the initial vertical velocity is zero, as the jumper starts from rest vertically. The acceleration due to gravity (g) is approximately 9.8m/s^2, pointing downward. The final vertical displacement (d) is 910m - 150m = 760m (distance from the top to 150m above the valley floor).
Using the equation of motion for vertical displacement:
d = (1/2) * g * t^2
Where:
d = displacement
g = acceleration due to gravity
t = time
Rearranging the equation, we get:
t = sqrt(2 * d / g)
Now, let's substitute the known values:
t = sqrt(2 * 760m / 9.8m/s^2)
t ≈ sqrt(155.1s^2) ≈ 12.5s
Therefore, the jumper is in freefall for approximately 12.5 seconds.
h=H-H(o) =910-150 =760 m/s
h=g•t^2/2,
t=sqrt(2•g•h) = sqrt(2•9.8•760)=122 s.
The horizontal component of the velocity doesn’t influence on the time of free fall.