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Question: You throw a stone vertically upward with an initial speed of 6.1 from a third-story office window.?
If the window is 13 above the ground, find the time the stone is in flight
AND
Find the speed of the stone just before it hits the ground

The answer in the following: (Hint: The answer is not necessarily.)

Answer by Jim
Numbers in physics without UNITS are nude and some people :>) might consider that rude.

Answer by SSJIV ~ Do u know CPR ? o_O
I assume that ur Measurements are in meters and meter per sec

Time taken for Velocity of Stone to become Zer0 =
V= u – at
0=6.1 -10t
t = 0.61 seconds
total time = 2*0.61 = 1.22 sec (this time is taken by stone to go to the highest peak and return back to the window on the third story building)
Distance traveled in 2 from movement —->
s=ut – 1/2at^2
=[6.1*.66 - 0.5(.66)^2] * 2 (multiply by two to calculate total distance ie from window to top and from top to window)
=3.8082 * 2 m
The total distance from the top most point of the Stone to the bottom is 13+ 3.8082/2 m
=14.9041m
At topmost point velocity = 0
At bottom velocity = v (let)
V^2=u^2+2aS
v^2=0 + 2*10*14.9041
V=17.26 m/s^2

Answer by The Enlightened
Assuming all units are in their respective SI form;
Speed = 6.1 m/s and distance = 30m.
At max height, velocity is zero…
So, v^2 = u^2 + 2as
0^2 = 6.1^2 + 2 x -9.8 x s
So, distance upwards ‘s’ = 6.1^2/19.6
s = 1.9 meters…
So distance above ground = 30 + 1.9
= 31.9 meters.
Velocity at impact, use the same equation with initial velocity zero (velocity at the top);
v^2 = u^2 + 2as
v^2 = 0^2 + 2 x 9.8 x 31.9
v^2 = 625.21
v = 25 m/s when reaching ground.
.
Time of flight;
Dispalcement is -30m (considering window level as zero)…
S = ut + (1/2)at^2
-30 = 6.1 x t + (1/2) x 9.8 x t^2
Solve to find ‘t’.

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