A squirrel is resting in a tall tree when it slips from a branch that is 43 m above the ground. It is a very agile squirrel and manages to land safely on another branch after only 0.47 s. What is the height of the branch it lands on?
Δh=gt²/2 = 9.8•0.47²/2 =1.08
h=H – Δh =43-1.08=41.92 m
To find the height of the branch the squirrel lands on, we need to use the equation of motion for free fall.
The equation is: h = ut + (1/2)gt^2
Where:
h = height (final position)
u = initial velocity (initially at rest, so u = 0)
g = acceleration due to gravity (approximately 9.8 m/s^2)
t = time taken to fall
In this case, the squirrel falls for 0.47 s.
Plugging in the values into the equation, we have:
h = 0(0.47) + (1/2)(9.8)(0.47)^2
Simplifying the equation:
h = 0 + 0.5(9.8)(0.22)
h = 0 + 1.078
h ≈ 1.08 m
Therefore, the height of the branch the squirrel lands on is approximately 1.08 meters.