Question

A 10 kg monkey swings on a vine from a point which is 40 m above the junglefloor to a point which is 15 m above the floor. If the monkey was moving at 2 m/sinitially, what will be his speed at the 15 m point?

Answer 1

In order to determine the final speed of the monkey, use the following equatio, for the final speed in a free fall motion:

`v^2=v^2_o+2gy`

where,

vo: initial speed = 2m/s

g: gravitational acceleration constant = 9.8m/s^2

y: vertical distance traveled = 40m - 15m = 25m

Replace the previous values of the parameters into the formula for v^2, and apply square root:

`\begin{gathered} v^2=(2(m)/(s))^2+2(9.8(m)/(s^2))(25m) \\ v^2=494(m^2)/(s^2) \\ v=\sqrt[]{494(m^2)/(s^2)} \\ v\approx22.23(m)/(s) \end{gathered}`

Hence, at the 15 m point the monkey will have a speed of approximately 22.23m/s