A skydiver jumps from an airplane at an altitude of 2,500 ft. He falls under the force of gravity until he opens his parachute at an altitude of 1,000 ft. Approximately how long does the jumper fall before he opens his chute?
the answer is chicken
He only falls 1500 ft before the chute opens. So, change d to 1500 in the above solution and then rework the numbers.
Gooday, after banging my head against the wall for 2 hours, not being able to find an explanation, I finally got the answer and I'm content to share my process. The equation utilized to solve this problem is h(t) = at^2 + vt + h(0) (h(t) is the height given the time, a is the acceleration due to gravity, v is initial velocity, and h(0) is the initial height)
The original problem from edgenuity shows an equation with a being 16ft/s 🥴. Acceleration due to gravity is 32ft/s on earth.
Our equation to solve for t looks like: -1500 = -32t^2 + 0v + (2500-1000)
h(t) is -1500 because we had a downwards (negative) displacement of 1500
0v parachuter started off with no velocity and just dropped
-32 <--- negative cuz we falling
Then: -1500 = -32t^2+1500
-3000 = -32t^2
93.75 = t^2
√(93.75) = √(t^2)
t = 9.682458... 😌👍
A skydiver jumps from an airplane at an altitude of 2,500 ft. He falls under the force of gravity until he opens his parachute at an altitude of 1,000 ft. Approximately how long does the jumper fall before he opens his chute?
To calculate the approximate time it takes for the skydiver to fall before opening the parachute, we can use the principles of physics. The distance fallen by an object in freefall can be determined using the equation:
d = 0.5 * g * t^2
where:
d is the distance fallen,
g is the acceleration due to gravity (32.2 ft/s^2), and
t is the time taken.
We can rearrange the equation to solve for t:
t = sqrt(2 * d / g)
To find the time, we need to calculate the distance fallen. The fall distance is the difference in altitudes between the starting point (2,500 ft) and the location where the parachute is opened (1,000 ft):
fall distance = 2,500 ft - 1,000 ft = 1,500 ft
Now we can substitute the values into the equation and calculate the time:
t = sqrt(2 * 1,500 ft / 32.2 ft/s^2)
Using a calculator, the approximate time is:
t ≈ 6.14 seconds
Therefore, the skydiver falls for approximately 6.14 seconds before opening his parachute.
d=0.5*g*t^2
therefore, t^2= d/ (0.5*g)
where:
d= 2500 ft
g = 32.2 ft /sec^2
solving t:
t^2 = 2500/ (0.5*32.2)
t^2 = 155.28 sec^2
t= 12.46 sec