A skydiver wants to jump out of a plane and be in free fall while she sketches some motion graphs for a Gr. 11 physics problem. She figures the quickest she can complete the sketches is 17 s. She then wants to open her parachute at the last moment so she can reach a safe velocity just when she reaches the ground. Generally, a skydiver can walk away unharmed when they hit the ground at 5.6 m/s. The average acceleration that the parachute can provide is 18.1 m/s2 [upwards].
How high will the plane be when she jumps? We can assume that she initially has no velocity in the vertical direction.
Well, if our skydiver wants to sketch some motion graphs, I hope she packed a lot of pencils! Anyway, let's get into the physics of this daring feat.
To find out how high the plane will be when she jumps, we can use the equation of motion for free fall. We know that the average acceleration provided by the parachute is 18.1 m/s² upwards, and we want to find the distance traveled. Let's call it 'd'.
The equation we can use is: d = (1/2) * a * t², where 'a' is the acceleration and 't' is the time in free fall.
Since our skydiver wants to complete the sketches in 17 seconds, we can plug in the values:
d = (1/2) * 18.1 m/s² * (17 s)²
Now, let me grab my calculator... *beep boop beep* Crunching the numbers... and voila! The height of the plane when she jumps will be approximately:
d = 2857.59 meters
That's quite a drop! Hopefully, she'll have enough time to finish all her sketches before she reaches the ground. And don't worry, she'll have a nice parachute to slow her down, so she'll land safely with a velocity of 5.6 m/s.
Just remember, if you're ever skydiving and sketching motion graphs at the same time, make sure to keep your pencils secure. You wouldn't want to lose them on the way down!