Respuesta :
Answer:
velocity is 13.1 m/s and at an angle of 90 degree with horizontal with respect to cart
Explanation:
As we know that cart speed is uniform and it is moving at 30 m/s
so here we can say that
[tex]v_x = 30 m/s[/tex]
now the cart is displaced by d = 80 m
so here the time taken by cart too move above distance
[tex]30 = \frac{d}{t}[/tex]
[tex]t = \frac{80}{30} = \frac{8}{3} s[/tex]
now in the same time the projectile must return to the cart so its displacement in y direction must be zero
[tex]\Delta y = v_y t + \frac{1}{2} at^2[/tex]
[tex]0 = v_y (\frac{8}{3}) - \frac{1}{2}(9.8)(\frac{8}{3})^2[/tex]
[tex]v_y = 13.1 m/s[/tex]
so here horizontal speed and vertical speed would be
[tex]v_x = 30 m/s[/tex]
[tex]v_y = 13.1 m/s[/tex]
here velocity with respect to cart is 13.1 m/s and it is at 90 degree with horizontal
Speed (relative to the cart) : [tex]\displaystyle v_x=30 ,v_y=\frac{40}{30}[/tex]
angle (to the horizontal) : θ = 23.962°
Further explanation
Projectile motion is a motion by particles that move along a curved path and are affected by gravity
The velocity in the horizontal direction is constant
The velocity in the vertical direction changes linearly (the acceleration is constant= g)
[tex]\displaystyle v_{x}=v_{0}\cos(\theta)[/tex]
[tex]{\displaystyle v_{y} = v_{0} \sin (\theta) -gt}[/tex]
• Horizontal and vertical displacement
[tex]x=v_{0} t \cos (\theta)[/tex]
[tex]{\displaystyle y=v_{0}t \sin (\theta) - {\frac {1} {2}} gt ^{2}}[/tex]
• Total time
[tex]{\displaystyle t={\frac {2v_{0} \sin (\theta)} {g}}}[/tex]
• Time to reach the maximum height (h):
[tex]{\displaystyle t_{h}={\frac {v_ {0} \sin (\theta)} {g}}}[/tex]
• The maximum height of projectile
[tex]\displaystyle h={\frac {v_{0}^{2} \sin^{2} (\theta)} {2g}}[/tex]
A cart is moving with constant speed of 30 m/s
so [tex]\displaystyle v_x=vo_x=30\frac{m}{s}[/tex]
so the total time taken by cart
[tex]\displaystyle t=\frac{x}{v} =\frac{80\:m}{30\frac{m}{s} } \\\\\displaystyle t=\frac{8}{3}s[/tex]
Time of flight = total time , then
[tex]{\displaystyle t={\frac {2v_{0} \sin (\theta)} {g}}}[/tex]
[tex]{\displaystyle \frac{8}{3} =\frac{2v_0sin\theta}{10}[/tex]
[tex]\displaystyle v_0sin\theta=\frac{40}{3}[/tex]
Because :
[tex]\displaystyle v_x=v_o\:cos\:\theta=30[/tex]
and
[tex]\displaystyle v_y=v_0\:sin\:{\theta}=\:\frac{40}{3}[/tex]
we can make tan θ =
[tex]\displaystyle\frac{vo\:sin\:\theta}{vo\:cos\:\theta}=\frac{\frac{40}{3} }{30}=\frac{4}{9}[/tex]
[tex]\displaystyle tan\:\theta=\frac{4}{9}[/tex]
θ = 23.962°
Learn more
velocity position vector
https://brainly.com/question/2005478
resultant velocity
https://brainly.com/question/4945130
The position of a training helicopter
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Keywords : a cart, projectile, speed, horizontal, vertical
