Respuesta :
Answer:
159.37 m
Explanation:
t = Time taken
u = Initial velocity
v = Final velocity
s = Displacement
a = Acceleration
g = Acceleration due to gravity = 9.81 m/s² = a
r = Radius = 12 m
Time taken to fall
[tex]s=ut+\frac{1}{2}at^2\\\Rightarrow 0.809=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{(20+12)\times 2}{9.81}}\\\Rightarrow t=2.554\ s[/tex]
Speed is given by
[tex]v=r\omega\\\Rightarrow v=12\times 5.2\\\Rightarrow v=62.4\ m/s[/tex]
Distance is given by
[tex]s=vt\\\Rightarrow s=62.4\times 2.554\\\Rightarrow s=159.37\ m[/tex]
The point P is 159.37 m from the base.
The projectile motion allows to find the result for the fall distance is:
- The fall distance is: x = 132.3 m
Given parameters.
- System height h = m
- Helix length l= 12 m
- Angular velocity w= 5.2 rad/s.
To find.
- The fall distance.
Projectile motion.
Projectile launch is an application of kinematics to motion in two dimensions, where on the x-axis there is no acceleration and on the y-axis the acceleration is the acceleration of gravity.
Let's look for the time that the body takes to reach the ground, since the body leaves horizontally, its initial vertical speed is zero.
y = y₉ + [tex]v_{oy}[/tex] t – ½ g t²
The height of the body when it reaches the ground is zero.
y= 0.
The initial height from where the body leaves is i = h+ l
y₀ = 20+ 12= 22m
we substitute
0 = y₀ + 0 – ½ g t²
t = [tex]\sqrt{\frac{2y_o}{g} } [/tex]
Let's calculate.
t= [tex]\sqrt{ \frac{2\ 22.0 }{9.8} }[/tex]
t = 2.12s
Linear and angular velocity are related.
v= rw
v= 12 5.2
v= 62.4 m/s
On the x-axis there is no acceleration so we can use the uniform motion relationship.
x= vt
Let's calculate.
x= 62.4 2.12
x = 132.3m
In conclusion using projectile motion we can find the result for the fall distance is:
- The fall distance is: x = 132.3 m
Learn more about parabolic motion here: brainly.com/question/24216590
