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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
Answer:
-10.0 cm
Explanation:
OK so in the equation they're having you use the variables are:
So we simply plug in the variables: