Question

Janet jumps off a high diving platform with

Answer 1

Answer:

a) $y_{o}=26m$

b)$x=6m$

Explanation:

From the exercise we know the initial horizontal velocity and the time that Janet takes to land in the water

$v_{ox}=2.47m/s\\t=2.3s\\g=-9.8m/s^2$

a) Knowing the formula for free falling objects, we can calculate how high is the platform

$y=y_{o}+v_{oy}t+\frac{1}{2}gt^2$

At t=2.3m the position of Janet is y=0m

$0=y_{o}-\frac{1}{2}(9.8m/s^2)(2.3s)^2$

$y_{o}=\frac{1}{2}(9.8m/s^2)(2.3s)^2=26m$

So, the platform is 26m tall

b) According to dynamics the displacement of an object can be analyze using the following formula:

$x=x_{o}+v_{ox}t+\frac{1}{2}a_{x} t^2$

$x=v_{ox}t=(2.47m/s)(2.3s)=6m$

So, Janet lands 6 m far from the base of the platform