Respuesta :
Answer:
a) x= 1618.3 m
b)V=180.59 m/s
Explanation:
Speed of plane = 489 Km/h
Speed of plane = 489 x 5/18 m/s = 135.33 m/s
Horizontal velocity of crate , u = 135.33 m/s
h= 715 m
a)
In vertical direction
[tex]h=v_ot+\dfrac{1}{2}gt^2[/tex]
The initial velocity in vertical direction =0 m/s
[tex]715=\dfrac{1}{2}\times 10\times t^2[/tex]
t = 11.95 s
So the distance travel ,x= u .t
x = 135. 33 x 11.95
x= 1618.3 m
b)
The speed of crate V
[tex]v^2=v_o^2+2gh[/tex]
[tex]V^2=2\times 10\times 715[/tex]
v=119.58 m/s
u = 135.33
So the resultant velocity V
[tex]V=\sqrt{v^2+u^2}[/tex]
[tex]V=\sqrt{119.58^2+135.33^2}[/tex]
V=180.59 m/s
The equation of motion is the relation between the distance, velocity, acceleration and time of a moving body.
- a) The horizontal distance traveled by crate is 1634 meters.
- b) The speed of the crate, in m/s, just before it hit the ground is 180.59 m/s.
What is the second equation of motion?
The equation of motion is the relation between the distance, velocity, acceleration and time of a moving body.
The second equation of the motion for distance can be given as,
[tex]s=ut+\dfrac{1}{2}at^2[/tex]
Here, [tex]u[/tex] is the initial body, [tex]a[/tex] is the acceleration of the body and [tex]t[/tex] is the time taken by it.
Given information-
The horizontal speed of the airplane is 489 km/h.
The altitude of the plane is 715 m.
Convert the horizontal speed of the airplane in meter per second by multiplying it with factor 5/18. Thus,
[tex]v_a=489\dfrac{5}{18} \\v_a=135.33\rm m/s[/tex]
- a) The horizontal distance traveled by crate is-
The distance traveled in the vertical direction is 715 m and velocity of the airplane is 135.33 m/s.
As the initial velocity is zero for vertical direction. Thus put the value in the distance formula to find the time,
[tex]715=0+\dfrac{1}{2}9.81\times t^2\\t=12.07\rm s[/tex]
Now the horizontal distance can be given as,
[tex]s=ut\\s=135.33\times12.07\\s=1634\rm m[/tex]
The horizontal distance traveled by crate is 1634 meters.
- b) The speed of the crate, in m/s, just before it hit the ground-
The initial velocity is zero for this case, thus use the velocity formula to find the speed of crate as,
[tex]v^2=(u)^2+2gs\\v^2=2\times 9.81\times715\\v=118.44 \rm m/s[/tex]
The speed of the crate, in m/s, just before it hit the ground is the resultant velocity of horizontal velocity and vertical velocity. thus,
[tex]V=\sqrt{135.33^2+119.58^2}\\V=180.59 \rm m/s[/tex]
Thu, the speed of the crate, in m/s, just before it hit the ground is 180.59 m/s.
Hence,
- a) The horizontal distance traveled by crate is 1634 meters.
- b) The speed of the crate, in m/s, just before it hit the ground is 180.59 m/s.
Learn more about the equation of motion here;
https://brainly.com/question/13763238
