This is a projectile motion problem, so, we use the formula for trajectory:
y =xtanα + gx^2/2v^2(cosα)^2
where
y is the vertical distance (y = 50 m)
x is the horizontal distance (x=90 m)
α is the angle of trajectory; since it levels of HORIZONTALLY, α = 0°
v is the initial velocity
g is the acceleration due to gravity which is 9.81 m/s^2
Substituting to the formula,
50 =90tan(0°) + (9.81)(90)^2/2v^2(cos0°)^2
v = 28.2 m/s
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
Approximately 100-120 wolves
Answer:
Simply put, slope refers to the steepness of a line. The larger the slope, the steeper the line. Slope is often referred to as 'rise over run' because it is calculated by the change in the vertical (rise) divided by the change in the horizontal (run). ... For example, a high value of slope means a very steep line.