A running mountain lion can make a leap 10.0 m long, reaching a maximum height of 3.0 m.?

a.What is the speed of the mountain lion just as it leaves the ground?

b.At what angle does it leave the ground?

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A running mountain lion can make a leap 10.0 m long, reaching a maximum height of 3.0 m.?

a.What is the speed of the mountain lion just as it leaves the ground?

b.At what angle does it leave the ground?

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When the lion is at the maximum height, its vertical velocity is 0 m/s. Let’s use the following equation to determine the initial vertical velocity.

vf^2 = vi^2 + 2 * g * d

0 = vi^2 + 2 * -9.8 * 3

vi = √58.8

This is approximately 7.7 m/s. This happens during one half of the total time. Let’s use the following equation to determine this time.

vf = vi – g * t

0 = √58.8 – 9.8 * t

t = √58.8 ÷ 9.8

This is approximately 0.78 seconds. The total time is twice this time. During the total time, the lion moves a horizontal distance of 10 meters. Let’s use the following equation to determine the horizontal component of the lion’s initial velocity.

d = v * t

10 = v * 2 * √58.8÷ 9.8

v = 49 ÷ √58.8

This is approximately 6.4 m/s.

Speed = √[58.8 + (49 ÷ √58.8)^2]

This is approximately 10 m/s. To determine the angle from horizontal, use the following equation.

Tan θ = Vertical ÷ Horizontal

Tan θ = √58.8 ÷ (49 ÷ √58.8) = 1.2

The angle is approximately 50˚. I hope this is helpful for you.

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10 = V^2/g*sin 2Θ

V*sin Θ = √19.612*3 = 7.670

V = 7.670/sin Θ

10*9.806 = 58.8360/(sin Θ*sin Θ)*2*sin Θ*cos Θ

98.06 = 58.836*2*cot Θ

cot Θ = 98.06/(58.836*2) = 0.833

tan Θ = 1/0.833 = 1.200

Θ = arctan 1.20 = 50.19°

V = 7.670/sin Θ = 7.67/0.767 = 10.0 m/sec

I think this is the correct answer for A running mountain lion can make a leap 10.0 m long, reaching a maximum height of 3.0 m.?

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From V^2 = U^2 - 2gH = 0 her vertical speed is Uy = sqrt(2gH). And from X = Ux t = Ux sqrt(2H/g) her horizontal speed is Ux = X/sqrt(2H/g).

a. U = sqrt(Uy^2 + Ux^2) = sqrt(2gH + X^2/(2H/g)) = sqrt(2*9.8*3 + 100/(2*3/9.8)) = 14.9 m/s. ANS.

b. theta = ATAN(Uy/Ux) = ATAN(sqrt(2gH)/X/sqrt(2H/g)) = ATAN(sqrt(2*9.8*3)/(10/sqrt(2*3/9.8))) = 29 deg re the horizontal. ANS.

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H = 3m = g*t^2/2 = 9.81*t^2/2 --> t = sqrt(3*2/9.81) = sqrt(0.612) = 0.782 s to get to the maximum height, and 0.782 s to come back down again.

In the vertical direction,

Vi = g*t = 7.67 m/s

In the horizontal direction, velocity is constant. V = 10m/1.564s = 6.39 m/s

The speed is sqrt(Vx^2 + Vy^2). The angle is atan(Vy/Vx).