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

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?

This is the Best Answer for A running mountain lion can make a leap 10.0 m long, reaching a maximum height of 3.0 m.?

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.

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.?

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.

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).