To rearrange these for v and r we must use BEDMAS.
First, times both sides by r, getting Ar = v²r/r
You can cancel out r/r to get Ar = v²
To get v, square root both sides. √Ar = √v²
Cancel out the square root and the squared v = √Ar
To get r, go back to Ar = v², then divide both sides by A. rA/A = v²/A
Then cancel out A/A to get r. r = v²/A
Answer:
a) v = 4.64 m / s
, b) t = 0.947 s
, c) t = 0.947 s
Explanation:
We will work on this exercise with vertical launch kinematics, let's start by calculating the height of the jumper in the SI system
y₀ = 5 ’(0.3048 m / 1’) + 7 ”(2.54 10-2 m / 1”) = 1.70 m
The distance they give is the height of the jump
y = 1.10 m
Let's use energy conservation
Starting point. On the floor
Em₀ = K = ½ m v²
Final point. Maximum height
Em_{f} = U = m g y
Em₀ =
½ m v² = m g y
v = √2gy
Let's calculate
v = √(2 9.8 1.10)
v = 4.64 m / s
b) Air time is the time to go up plus the time to go down, which is the same
For maximum height the speed is zero
v = v₀ - g t₁
t₁ = v₀ / g
t₁ = 4.64 /9.8
t₁ = 0.4735 s
The total time is
t = 2 t₁
t = 2 0.4735
t = 0.947 s
c) if it takes a distance of 0.40 to reach speed, what is the acceleration, as it stands on the floor its initial speed is zero
v² = v₀² + 2 a x
a = v² / 2x
a = 4.64²/2 0.40
a = 26.9 m / s²