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Sagot :
I thought you were going to ask how long it takes for the stone to hit
the canyon floor at the bottom of the cliff. But you're not asking for
anything after the stone barely misses the edge of the cliff on its way
down, and you're only asking about things that happen above the
edge of the cliff. So we don't need to know anything about the cliff,
or its height above anything else. The thrower might just as well be
standing in his front yard when he throws the stone straight up.
-- Gravity robs 9.8 m/s from the stone's upward speed every second.
So the stone reaches its maximum height, runs out of upward gas, and
starts falling, in (20 / 9.8) = 2.04 seconds after the toss.
-- Its average speed during that time is (1/2) (20 + 0) = 10 m/s .
In 2.04 seconds at that average speed, it rises (2.04 x 10) = 20.4 meters .
The time taken by the stone to reach the maximum height above the cliff is [tex]\boxed{2.04\text{ s}}[/tex] and the maximum height reached by the stone above the cliff is [tex]\boxed{20.4\text{ m}}[/tex] .
Explanation:
Given:
The initial velocity of the stone is [tex]20\text{ m/s}[/tex].
The height of the cliff is [tex]100\text{ m}[/tex].
Concept:
As the stone is thrown upward from the top of the cliff, the stone will move under the action of the acceleration due to gravity. The speed of the stone will decrease as it moves up because the acceleration on the stone is acting in the direction opposite to its motion.
The speed of the stone at its maximum height will be zero where is comes to rest for a moment.
The time taken by the stone to reach the maximum height is given by:
[tex]\boxed{v_f=v_i-gt}[/tex]
Here, [tex]v_f[/tex] is the final velocity, [tex]v_i[/tex] is the initial velocity, [tex]g[/tex] is the acceleration and [tex]t[/tex] is the time taken by the stone.
Substitute the values in above expression.
[tex]\begin{aligned}0&=20-(9.81)t\\t&=\dfrac{20}{9.81}\text{ s}\\&=2.04\text{ s}\end{aligned}[/tex]
The distance covered by the stone in reaching the maximum height is given by:
[tex]\boxed{v_f^2=v_i^2-2gS}[/tex]
Here, [tex]S[/tex] is the distance covered by the stone.
Substitute the values in above expression.
[tex]\begin{aligned}(0)^2&=(20)^2-2(9.81)S\\S&=\dfrac{400}{19.62}\text{ m}\\&=20.4\text{ m}\end{aligned}[/tex]
Thus, The time taken by the stone to reach the maximum height above the cliff is [tex]\boxed{2.04\text{ s}}[/tex] and the maximum height reached by the stone above the cliff is [tex]\boxed{20.4\text{ m}}[/tex] .
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Answer Details:
Grade: Senior School
Subject: Physics
Chapter: Kinematics
Keywords:
stone, initial velocity, final, thrown, cliff, canyon floor, maximum height, zero, acceleration, gravity, time taken, distance.
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