Question

In the standing broad jump, one squats and then pushes off with the legs to see how far one can jump. Suppose the extension of the legs from the crouch position is 0.600 m and the acceleration achieved from this position is 1.25 times the acceleration due to gravity, g. How far can they jump? State your assumptions. (Increased range can be achieved by swinging the arms in the direction of the jump.)

Answer 1

Answer:

$H_{max} = 0.75 m$

Explanation:

Acceleration achieved is given as

a = 1.25 g

So we will have

$a = 12.26 m/s^2$

now the velocity that he will gain till he moved 0.600 m is given as

$v^2 - v_i^2 = 2 a d$

$v^2 - 0 = 2(12.26)(0.600)$

$v = 3.84 m/s$

now with this speed the maximum height achieved by him

$H_{max} = \frac{v_f^2 - v_i^2}{2g}$

$H_{max} = \frac{0 - 3.84^2}{2(-9.81)}$

$H_{max} = 0.75 m$

Answer 2

⇒Answer:

In the standing broad jump, one squats and then pushes off with the legs to see how far one can jump. Suppose the extension of the legs from the crouch position is 0.600 m and the acceleration achieved from this position is 1.25 times the acceleration due to gravity, g . How far can they jump? State your assumptions. (Increased range can be achieved by swinging the arms in the direction of the jump.)

Step-by-Step Solution:

Solution 35PE

This question discusses about the increased range. So, we shall assume that the angle of jumping will be  as the horizontal range is maximum at this angle.

Step 1 of 3<

/p>

The legs have an extension of 0.600 m in the crouch position.

So,  m

The person is at rest initially, so the initial velocity will be zero.

The acceleration is  m/s2

Acceleration  m/s2

Let the final velocity be .

Step 2 of 3<

/p>

Substitute the above given values in the kinematic equation  ,

m/s

Therefore, the final velocity or jumping speed is  m/s

Explanation: