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

Answer 1

Answer:

The distance up to which they can jump is 1.5 m

Solution:

As per the question:

The extension of the legs, d = 0.600 m

Acceleration, $a = 1.25\times g$

where

g = acceleration due to gravity = $9.8 m/s^{2}$

Now,

Let the initial velocity of the jump be v m/s

So, by the third eqn of motion:

$v'^{2} = v^{2} + 2ad$            (1)

where

v' = final velocity

a = acceleration

d = distance between the legs

Also, for maximum range, angle, $\theta = 45^{\circ}$

$x = \frac{v'^{2}sin2\theta}{g}$     (2)

x = maximum horizontal range

Thus, using eqn (1):

$v'^{2} = 0^{2} + 2\times 1.25\times 9.8\times 0.600$

$v' = \sqrt(14.7) = 3.834 m/s$

Now, using eqn (2):

$x = \frac{14.7sin2(45^{\circ})}{9.8} = 1.5 m$