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
vo = initial velocity of lion
=
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 Sin
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 Sin
= 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 Sin
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 Sin)
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
Cos
a = acceleration = 0
X = displacement = 10
t = time of travel = 1.6
using the equation
X = Vox t + (0.5) at2
10 = (Vo Cos
) (1.6)+ (0.5) (0)(1.6)2
Vo Cos
= 6.3 m/s eq-2
dividing eq-1 by eq-2
Vo Sin/(Vo
Cos)
= 7.7/6.3
tan
= 1.22
= tan-1(1.22)
= 50.7 deg
using eq-1
Vo Sin
= 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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