Jumping Lab Question ( Kinematics )

[limework]

Homework Statement

a.Standing(Tip Toes) - 0.15m
b.Crouching(As before you jump) - 0.35m
c.At the height of a jump - 0.36m
d.At the bottom of your landing - 0.67

Mass = 75kg

1. Find the initial velocity at "take off" ( remember to use magnitude and direction of your acceleration while you are in the air- use c --> d

2. Find the magnitude and the direction of the average acceleration ( Use b--->c)

3. Find the magnitude and direction of the average acceleration from when you "Touched down" to when you stopped in the crouch position ( use e---> f)

Homework Equations

The Attempt at a Solution

I am new to Physics and i asked my teacher for help but he said he would not answer any questions on this lab?

Can you please help. Maybe explain to me how to do it but I really want to learn.

Thanks!

 

[limework]

Anyone?/

 

[nlightner]

Hello,

I am happy to help you with your lab question. First, let's define some variables:
- Initial velocity (u) = ?
- Final velocity (v) = ?
- Acceleration (a) = ?
- Distance (d) = 0.36m
- Mass (m) = 75kg

1. To find the initial velocity at take off, we can use the equation v^2 = u^2 + 2ad, where v is the final velocity, u is the initial velocity, a is the acceleration, and d is the distance. We know the final velocity is 0 at the peak of the jump and the distance is 0.36m. So, let's plug in the values we know: 0^2 = u^2 + 2(0.36)(-9.8). Solving for u, we get u = 1.9 m/s. Therefore, the initial velocity at take off was 1.9 m/s.

2. To find the average acceleration from b to c, we can use the equation a = (v-u)/t, where v is the final velocity, u is the initial velocity, and t is the time. We know the initial velocity is 0.35 m/s (from crouching) and the final velocity is 0 m/s (at the peak of the jump). We also know that the time taken to go from b to c is the same as the time taken to go from c to d. So, we can use the equation v = u + at to find the time (t). Plugging in the values, we get 0 = 0.35 + a(t). Solving for t, we get t = -0.35/a. Now, let's substitute this value of t in the first equation to find the average acceleration: a = (0-0.35)/(-0.35/a) = 0.35a/0.35 = a. Therefore, the magnitude and direction of the average acceleration from b to c is a = -9.8 m/s^2 (acceleration due to gravity) downwards.

3. To find the average acceleration from e to f, we can use the same equation as in part 2. Now, the initial velocity is 0 m/s (from landing) and the final velocity is 0.35 m/s (in the