Question

An acorn is dropped and reaches the ground in 2.0s. What is the acorn’s velocity just as it hits the ground? How high above the ground was it released?

Answer 1

Take into account that the vertical distance traveled by the acorn can be calculated by using:

`h=(1)/(2)gt^2`

where,

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

t: time acorn takes to hit the ground = 2.0s

h: height at which acorn is initially = ?

Replace the given values of the parameters into the previous formula and simplfiy:

`h=(1)/(2)(9.8(m)/(s^2))(2.0s)^2=19.6m`

Hence, the acorn was released from a high of 19.6m above the ground.

The velocity just after acorn hits the ground is given by:

`\begin{gathered} v=gt \\ v=(9.8(m)/(s^2))(2.0s)=19.6(m)/(s) \end{gathered}`

Hence, the speed is 19.6 m/s