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.

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

2. At what angle does it leave the ground?

Thank you!

Answer

0 Maximum Height Motion alongo Y-axis ベ バ~、 Motion along X-axis om

vo = initial velocity of lion

theta=
angle of launch

consider the motion along the vertical direction or Y-direction
from initial position to the point of maximum height.

Voy = initial velocity of lion In Y-direction =
Vo Sintheta

a = acceleration = – 9.8 m/s2

Ymax = maximum height gained = 3 m

Vfy = final velocity in Y-direction at the maximum
height = 0 m/s

using the equation

Vfy2 = Voy2
+ 2 a Ymax

inserting the values

02 = Voy2 + 2 (- 9.8) (3)

Voy2 = 58.8

Voy = sqrt(58.8)

Voy = 7.7 m/s

Vo Sintheta
= 7.7 m/s eq-1

consider the motion along the vertical direction or
Y-direction from initial to final position.

Voy = initial velocity of lion In Y-direction =
Vo Sintheta

a = acceleration = – 9.8 m/s2

Y = displacement = 0 m

t = time of travel

using the equation

Y = Voy t + (0.5) at2

0 = (Vo Sintheta)
t + (0.5) (- 9.8) t2

using eq-1

0 = (7.7) t + (0.5) (- 9.8) t2

t = 1.6 sec

consider the motion along the horizontal direction or
X-direction

Vox = initial velocity In X-direction = Vo
Costheta

a = acceleration = 0

X = displacement = 10

t = time of travel = 1.6

using the equation

X = Vox t + (0.5) at2

10 = (Vo Costheta
) (1.6)+ (0.5) (0)(1.6)2

Vo Costheta
= 6.3 m/s eq-2

dividing eq-1 by eq-2

Vo Sintheta/(Vo
Costheta)
= 7.7/6.3

tantheta
= 1.22

theta
= tan-1(1.22)

theta
= 50.7 deg

using eq-1

Vo Sintheta
= 7.7

inserting the value of the angle

Vo Sin50.7 = 7.7

Vo = 9.95 m/s

0 Maximum Height Motion alongo Y-axis ベ バ~、 Motion along X-axis om

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