If a 54 pound boy falls off the top of the grand canyon which is 5996 feet deep, how many seconds will it take for him to hit the bottom?
his weight is not a factor
the free fall equation is ... h = -1/2 g t^2 + Vo t + Ho
... h is height , g is gravitational acceleration
... Vo is initial velocity , Ho is initial height
h = 0 at the bottom , g = 32 ft/s^2
0 = -16 t^2 + 0 t + 5996
solve the quadratic for t
0.5g*t^2 = 5996.
16t^2 = 5996,
To determine the time it takes for the boy to hit the bottom of the Grand Canyon, we can use the formula for free fall motion:
h = (1/2) * g * t^2
Where:
- h is the distance fallen (5996 feet in this case)
- g is the acceleration due to gravity (approximated as 32.2 feet/second^2)
- t is the time it takes for the object to fall (what we are trying to find)
Rearranging the formula to solve for t:
t = sqrt((2 * h) / g)
Now we can substitute the given values:
h = 5996 feet
g ≈ 32.2 feet/second^2
t = sqrt((2 * 5996) / 32.2)
Calculating this equation:
t ≈ sqrt(117.31)
Therefore, it will take approximately 10.83 seconds for the boy to hit the bottom of the Grand Canyon.