A mountain lion jumps to a height of 3.25 m when leaving the ground at an angle of 43.2°. What is its initial speed (in m/s) as it leaves the ground?

where [tex]v_i[/tex] and [tex]v_f[/tex] are the lion's initial and final vertical velocities, [tex]a[/tex] is its acceleration, and [tex]\Delta y[/tex] is the vertical displacement.

At its maximum height, the lion has 0 vertical velocity, so we have

[tex]0={v_i}^2-2gy_{\rm max}[/tex]

where g is the acceleration due to gravity, 9.80 m/s², and we take the starting position of the lion on the ground to be the origin so that [tex]\Delta y=y_{\rm max}-0=y_{\rm max}[/tex].

Let v denote the initial speed of the jump. Then

[tex]v_i=v\sin(43.2^\circ)=\sqrt{2\left(9.80\dfrac{\rm m}{\mathrm s^2}\right)(3.25\,\mathrm m)}\implies\boxed{v\approx11.7\dfrac{\rm m}{\rm s}}[/tex]

Cricetus Cricetus · Answer 2 · 2022-02-11T14:28:01+01:00

The initial speed will be "11.7 m/s".

Speed and Displacement:

Given:

Height = 3.25 m
Angle = 43.2°
Acceleration due to gravity = 9.8 m/s²

We know,

→ [tex]vf^2=v_i^2+2a\Delta y[/tex]

At max. height, vertical height be zero, then

→ [tex]0 = v_i^2-2gy_{max}[/tex]

or,

→ [tex]\Delta y = y_{max} -0 = y_{max}[/tex]

now,

The initial speed,

→ [tex]v_i = v \ Sin(43.2^{\circ})[/tex]

By substituting the values,

[tex]= \sqrt{1\times 9.8\times 3.25}[/tex]

[tex]= 11.7 \ m/s[/tex]

Thus the solution above is correct.

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