A squirrel is resting in a tall tree when it slips from a branch that is 40 m above the ground. It is a very agile squirrel and manages to land safely on another branch after only 0.71 s. What is the height of the branch it lands on?
To find the height of the branch the squirrel lands on, we can use the equations of motion. The key equation we will use is:
h = ut + (1/2)gt^2
Where:
h = height (unknown)
u = initial velocity (0 m/s as the squirrel starts from rest)
t = time taken (0.71 s)
g = acceleration due to gravity (-9.8 m/s^2, assuming downwards as negative)
Since the initial velocity is 0 m/s, the equation simplifies to:
h = (1/2)gt^2
Substituting in the given values:
h = (1/2)(-9.8 m/s^2)(0.71 s)^2
h = (1/2)(-9.8 m/s^2)(0.5041 s^2)
h = -4.9 m/s^2 * 0.5041 s^2
h = -2.46509 m
The negative sign indicates that the branch is below the starting point, but since height cannot be negative, we ignore the negative sign and take the absolute value.
The height of the branch the squirrel lands on is approximately 2.47 meters.