Answer :
Answer:
a) [tex]y_{o}=26m[/tex]
b)[tex]x=6m[/tex]
Explanation:
From the exercise we know the initial horizontal velocity and the time that Janet takes to land in the water
[tex]v_{ox}=2.47m/s\\t=2.3s\\g=-9.8m/s^2[/tex]
a) Knowing the formula for free falling objects, we can calculate how high is the platform
[tex]y=y_{o}+v_{oy}t+\frac{1}{2}gt^2[/tex]
At t=2.3m the position of Janet is y=0m
[tex]0=y_{o}-\frac{1}{2}(9.8m/s^2)(2.3s)^2[/tex]
[tex]y_{o}=\frac{1}{2}(9.8m/s^2)(2.3s)^2=26m[/tex]
So, the platform is 26m tall
b) According to dynamics the displacement of an object can be analyze using the following formula:
[tex]x=x_{o}+v_{ox}t+\frac{1}{2}a_{x} t^2[/tex]
[tex]x=v_{ox}t=(2.47m/s)(2.3s)=6m[/tex]
So, Janet lands 6 m far from the base of the platform
Answer:
Janet jumps horizontally off a high diving platform with a velocity of 2.76 m/s and lands in the water 2.3 s later. How high is the platform? The acceleration of gravity is 9.8 m/s. Answer in units of m.
How far from the base of the platform does she land? Answer and units of m.
Explanation: