Question

a firefighter dives at 7.5 m/sec from window to net. If window is 34 m above net, what speed does firefighter hit net. ACCELRATION IS 9.8 m/sec^2

Answer 1

Answer:

Step-by-step explanation:

This is more physics than math, but since "The book of nature is written in the language of mathematics"...

Anyway, the one-dimesional motion equation you want to use here resembles the quadratic we use to model parabolic motion. It is

$v^2=v_0^2+2a$Δx

where v is the final velocity of the firefighter, v₀ is the initial velocity of the firefighter, a is -9.8 m/s/s (negative because the motion is in the downward direction), and Δx is -34 m. This is also negative because where the firefighter lands is below the point at which he started. This is important, because if you leave the 34 as positive, you end up with a negative radicand...and that's not gonna work. Filling in the formula:

$v^2=(7.5)^2+2(-9.8)(-34)$

Paying close attention to significant digits here is important (at least it is when I teach physics in school!). Since there are 2 significant digits in 7.5, we need 2 in the product. 7.5 squared is 56.25 which will round to 56. Then on the right side of the plus sign, we need 2 significant digits as well since the number 2 is part of the equation. We don't count it as significant if it is part of the equation. 2(-9.8)(-34) = 666.4 but that will round to 670. NOW, since the rules for significant digits are different in multiplication and addition, you have to round each first, then take the square root of the sum, rounding at the end. In this case we will round to the place that holds the least significance between the 2 numbers. In our case that is the tens place, since the tens place is less significant than is the 1's place. The number 7 is in the tens place. (Following that rule is super tricky, and also quite maddening!)

What we have here is

$v^2=56+670$ and

$v=\sqrt{56+670}$

That leaves us with

v = 26.94438j

But don't forget we round that to the tens place which comes out in the end, finally, to

v = 30 m/s