Acougar jumps to a height of 3.10 m when leaving the ground at an angle of 40.7°. what is its initial speed (in m/s) as it leaves the ground?

Vo= 11.9595m/s

Explanation:

First we decompose the analysis on the horizontal X axis and the vertical Y axis. On the Y axis we know that the space traveled is the height, 3.1m, and the final speed is zero, when we reach the maximum height we can therefore establish these equations:

Dy= Voy×t - (1/2)×g×t²

Voy = component of the initial velocity on the Y axis.

t = time to reach maximum height.

g = acceleration of gravity.

Vfy= Voy - g×t

Vfy = component of the final velocity on the Y axis, maxium heigh.

Vfy= 0= Voy - 9.81(m/s²)×t ⇒ Voy= 9.81(m/s²)×t

This clearance I place in the first equation to clear the time:

3.1m= g×t² - (1/2)×g×t² = (1/2)×g×t²⇒ t= √(0.6320s²) = 0.795s

I place this time in the second equation to obtain the initial velocity component on the Y axis:

Voy= g×t= 9.81(m/s²)×0.795s= 7.7988m/s

Knowing the jump angle, 40.7 °, I use the Pythagorean theorem to determine the initial velocity, knowing the angle and the component of the initial velocity on the Y axis, being this one of the legs of the velocity triangle:

Voy= Vo×sen(40.7°)⇒ Vo= ( Voy/sen(40.7°) ) = (7.7988m/s) / sen(40.7°) ⇒

⇒Vo= 11.9595m/s

This is the initial speed of the cougar leaving the ground in its jump.

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