please tell me if these are correct
A track star in the long jump goes into the jump at 12 m/s and launches herself at 20.0 deg above the horizontal. What is the magnitude of her horizontal displacement?
(Assume no air resistance and that Ay= -g = -9.81 m/s^2)
i got 9.2 m
There are 6 books in a stack, and each book weighs 5 N. The coefficient of static friction between the books is 0.2. With what horizontal force must one push to start sliding the top 5 books off the bottom one?
i got 3N
A rock is thrown straight upward with an initial velocity of 9.6 m/s in a location where the acceleration due to gravity has a magnitude of 9.81 m/s^2. To what height does it rise?
i got 4.7
can you please show me how to solve for this last one
thanks
To solve for the height the rock rises, we can use the equations of motion.
The first thing we need to know is the time it takes for the rock to reach its maximum height. We can use the equation:
vf = vi + at
where vf is the final velocity (which is 0 m/s at the maximum height), vi is the initial velocity (9.6 m/s upwards), a is the acceleration (-9.81 m/s^2), and t is the time.
0 = 9.6 - 9.81t
Rearranging the equation, we get:
t = 9.6 / 9.81
t ≈ 0.98 seconds
Now we can use this value of t and the equation for displacement in vertical motion:
Δy = vit + (1/2)at^2
where Δy is the displacement in the vertical direction, vi is the initial velocity, t is the time, and a is the acceleration.
Δy = (9.6)(0.98) + (1/2)(-9.81)(0.98)^2
Δy ≈ 4.7 meters
Therefore, the rock rises to a height of approximately 4.7 meters.
To find the height to which a rock rises when thrown straight upward, you can use the kinematic equations of motion.
The equation that relates final velocity (vf), initial velocity (vi), acceleration (a), and displacement (d) for an object undergoing constant acceleration is:
vf^2 = vi^2 + 2ad
In this case, the final velocity at the top of the trajectory will be zero since the rock momentarily comes to a stop before falling back down. The initial velocity is 9.6 m/s, and the acceleration due to gravity is -9.81 m/s^2 (negative because it is acting opposite to the direction of the initial velocity).
Let's solve for displacement (d) when vf becomes 0:
0 = (9.6 m/s)^2 + 2(-9.81 m/s^2)d
Simplifying the equation, we have:
0 = 92.16 m^2/s^2 - 19.62 m/s^2*d
Rearranging the equation:
19.62 m/s^2*d = 92.16 m^2/s^2
d = 92.16 m^2/s^2 / 19.62 m/s^2
d ≈ 4.7 meters
So, the height to which the rock rises is approximately 4.7 meters. Therefore, your answer of 4.7 is correct.