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You throw a stone vertically upward with an initial speed of 6.0m/s from a third-story office window. If the window is 12m above the ground, find(a) the time the stone is in flight(b)the speed of the stone just before it hits the ground

Question

You throw a stone vertically upward with an initial speed of 6.0m/s from a third-story office window. If the window is 12m above the ground, find

(a) the time the stone is in flight

(b)the speed of the stone just before it hits the ground 

check_circleAnswer
Step 1

Given information:

Hight of the window (h) = 12 m.

Initial velocity (u) = 6 m/s

Let us consider the upward direction as +ve and down ward direction as -ve

We know that the object moves under acceleration due to gravity hence (a) = -g

Step 2

Consider the second equation of motion:

Consider the second equation of motion:
s ut
at2
Substitute the values according to the sign convention:
-글(9.81)t2
-12 6t
Re-arranging the above equation we get:
4.9t2 6t 12 = 0
Solving the above equation, we get
t 2.29 s (or) -1.067 s
since time can not be -ve, the time of flight of stone is 2.29 s.
help_outline

Image Transcriptionclose

Consider the second equation of motion: s ut at2 Substitute the values according to the sign convention: -글(9.81)t2 -12 6t Re-arranging the above equation we get: 4.9t2 6t 12 = 0 Solving the above equation, we get t 2.29 s (or) -1.067 s since time can not be -ve, the time of flight of stone is 2.29 s.

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Step 3

consider first equati...

V=u + at
Substitute the values according to the sign convention:
6-9.81(2.29) = -16.46
v
-ve sign indicates downward direction hence the final velocity of the stone before hitting the
ground is
v = 16.46 m/s downward
help_outline

Image Transcriptionclose

V=u + at Substitute the values according to the sign convention: 6-9.81(2.29) = -16.46 v -ve sign indicates downward direction hence the final velocity of the stone before hitting the ground is v = 16.46 m/s downward

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