A tiger leaps horizontally from a 5.9 meter high rock with a speed of 3.7 m/s.
Questions: How far from the base of the rock will the tiger land?
H=gt²/2 => t =sqrt(2H/g) =sqrt(2•5.9/9.8) =1.1 s.
X =v•t =3.7•1.1 = 4.07 m.
y=1/2at^2|
5.9=1/2(9.8)t^2|
1.204=t^2|
t=1.097s|
___________
x=t[(Vi+Vf)/2]
x=1.097[(3.7+0)/2]
x=2.03m
To find the distance from the base of the rock where the tiger will land, we can use the kinematic equation of motion. The equation is:
d = v * t + 0.5 * a * t^2
Where:
d = distance
v = initial velocity
t = time
a = acceleration
In this case, the initial velocity is the horizontal speed of the tiger, which is 3.7 m/s. Since the tiger leaps horizontally, there is no vertical acceleration. Therefore, we can assume the acceleration (a) to be zero.
Now, let's calculate the time it takes for the tiger to reach the ground. We will use the equation:
h = 0.5 * g * t^2
Where h = height and g = acceleration due to gravity (9.8 m/s^2).
Plugging in the given values, we have:
5.9 = 0.5 * 9.8 * t^2
Simplifying the equation, we get:
t^2 = (2 * 5.9) / 9.8
t^2 = 1.198
Taking the square root of both sides, we find:
t = √1.198
t ≈ 1.093
Now, we can substitute this value of t into the first equation to find the distance (d):
d = 3.7 * 1.093 + 0.5 * 0 * 1.093^2
d ≈ 4.044
Therefore, the tiger will land approximately 4.044 meters from the base of the rock.