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, [tex]a = 1.25\times g[/tex]
where
g = acceleration due to gravity = [tex]9.8 m/s^{2}[/tex]
Now,
Let the initial velocity of the jump be v m/s
So, by the third eqn of motion:
[tex]v'^{2} = v^{2} + 2ad[/tex] (1)
where
v' = final velocity
a = acceleration
d = distance between the legs
Also, for maximum range, angle, [tex]\theta = 45^{\circ}[/tex]
[tex]x = \frac{v'^{2}sin2\theta}{g}[/tex] (2)
x = maximum horizontal range
Thus, using eqn (1):
[tex]v'^{2} = 0^{2} + 2\times 1.25\times 9.8\times 0.600[/tex]
[tex]v' = \sqrt(14.7) = 3.834 m/s[/tex]
Now, using eqn (2):
[tex]x = \frac{14.7sin2(45^{\circ})}{9.8} = 1.5 m[/tex]
