A lemming runs horizontally off the side of a 27.8 m cliff at 1.12 m/s. How long will it take for the lemming to hit the water? How far will it travel in the horizontal direction?
u = 1.12 forever
Vi =initial vertical velocity = 0
v = Vi - 9.81 t
h = Hi + Vi t - 4.9 t^2
h = 27.8 - 4.9 t^2
0 = 27.8 - 4.9 t^2
so
t = sqrt (27.8/4.9)
use that t to get horizontal range
x = 1.12 t
To find the time it takes for the lemming to hit the water, we can use the equation of motion for vertical motion:
y = y0 + v0y*t + (1/2) * g * t^2
Where:
- y is the vertical displacement (27.8 m, given)
- y0 is the initial vertical position (0 m, since the lemming starts at the top of the cliff)
- v0y is the initial vertical velocity (0 m/s, since the lemming runs horizontally)
- g is the acceleration due to gravity (-9.8 m/s^2)
Plugging in the values, the equation becomes:
27.8 = 0 + 0*t + (1/2) * (-9.8) * t^2
Simplifying the equation gives:
4.9t^2 = 27.8
Dividing both sides by 4.9 gives:
t^2 = 5.673
Taking the square root of both sides gives:
t ≈ 2.38 seconds
Therefore, it will take approximately 2.38 seconds for the lemming to hit the water.
Now, to find how far the lemming will travel in the horizontal direction, we can use the equation of motion for horizontal motion:
x = x0 + v0x * t
Where:
- x is the horizontal displacement (unknown)
- x0 is the initial horizontal position (0 m, since the lemming starts at the edge of the cliff)
- v0x is the initial horizontal velocity (1.12 m/s, given)
- t is the time taken (2.38 seconds, calculated above)
Plugging in the values, the equation becomes:
x = 0 + 1.12 * 2.38
Simplifying the equation gives:
x = 2.6664 meters (rounded to four decimal places)
Therefore, the lemming will travel approximately 2.6664 meters in the horizontal direction.