A tiger leaps horizontally from a 6.5m high rock with a speed of 4 m/s how far from the base of the rock will she land?
0.5g*T^2 = 6.5.
4.9T^2 = 6.5
T = 1.15 s.
d = V*T = 4*1.15 =
What is 0.5g ?
0.5*9.8 = 4.9 m/s^2.
To find out how far from the base of the rock the tiger will land, we can use the kinematic equation:
s = ut + (1/2)at^2
Where:
s = final displacement (distance from the base of the rock)
u = initial velocity = 4 m/s
t = time taken to reach the ground
a = acceleration due to gravity = 9.8 m/s^2
First, let's find the time taken by the tiger to reach the ground. The vertical distance traveled can be calculated using the equation:
s = ut + (1/2)at^2
Since the tiger leaps horizontally, there is no initial vertical velocity (u = 0), and the equation simplifies to:
s = (1/2)at^2
Rearranging the equation:
2s/a = t^2
t = √(2s/a)
Substituting the values:
s = 6.5 m
a = 9.8 m/s^2
t = √(2 * 6.5 / 9.8)
t ≈ 0.9 seconds (rounded to one decimal place)
Now that we know the time, we can find the horizontal distance traveled using the equation:
s = ut
Since the initial horizontal velocity (u) is constant at 4 m/s, and the time taken (t) is 0.9 seconds:
s = 4 * 0.9
s ≈ 3.6 meters (rounded to one decimal place)
Therefore, the tiger will land approximately 3.6 meters from the base of the rock.