Respuesta :
Answer:
[tex]v = 1.97 m/s[/tex]
Explanation:
Time required by the ball to reach the bottom of the track is given as
[tex]y = \frac{1}{2}gt^2[/tex]
[tex]4.1 = \frac{1}{2}(9.81)t^2[/tex]
[tex]t = 0.91 s[/tex]
now the speed of the ball is given as
[tex]v = \frac{d}{t}[/tex]
here we have
d = 1.8 m
so we have
[tex]v = \frac{1.8}{0.91}[/tex]
[tex]v = 1.97 m/s[/tex]
The equation of motion is the relation between the distance, velocity, acceleration and time of a moving body.
The speed of the block when it leaves the track is 1.97 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 mass of the block is 472 g.
The initial distance of the block from the ground is 4.1 m.
The distance traveled by the block is 1.8 m.
The initial distance of the block is 4.1 and initial speed of the block is zero.
As it is falling only due to the force of acceleration of gravity. Thus the distance formula for the case can be given as,
[tex]\begin{aligned}4.1&=0+\dfrac{1}{2}\times9.81\times t^2\\t&=0.91 \em s\\\end[/tex]
The speed of the block is the ratio of distance traveled by it and the time taken by the block. Thus,
[tex]v_i=\dfrac{1.8}{0.91}\\v_i=1.97 \rm m/s[/tex]
Hence, the speed of the block when it leaves the track is 1.97 m/s.
Learn more about the equation of motion here;
https://brainly.com/question/13763238
