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-Dominant- [34]
3 years ago
9

A projectile is launched horizontally off a cliff. If the projectile is launched with more velocity, the time it spends in the a

ir will ____________________.
Physics
1 answer:
Sedaia [141]3 years ago
5 0

Answer:

If the projectile is launched horizontally the time it remains in the air will remain constant even if we increase the horizontal velocity.

Explanation:

The vertical displacement will be the same irrespective of the horizontal velocity assuming the projectile is launched from same height. The formula for vertical displacement is given by:

y = v_{oy} - 1/2gt^{2}

where, y = vertical displacement, t = time,

g = acceleration due to gravity,

v_{oy} = Initial vertical velocity

Since, projectile is launched horizontally, therefore v_{oy} = 0

so the equation becomes

y = - 1/2gt^{2}

The above equation shows that time spend in air is not dependent on horizontal launched velocity as there is no horizontal velocity in the formula therefore the time in air will remain constant even if we increase the horizontal velocity.

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Answer:

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From the question we are told that

    The length of the wire is L  = 3 \ m

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Since the wire are of the same material Young's modulus(Y)  is constant

So we have  

              \frac{F * L }{r^2 e}  =  \pi * Y = constant

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Now the ration between the first and the second wire is

         \frac{F_1}{F_2}  * \frac{L_1}{L_2} =  \frac{r*2_1}{r^2}  *  \frac{e_1}{e_2}

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   We have

           \frac{L_1}{L_2} =   \frac{e_1}{e_2}

substituting values

          \frac{3}{6} =   \frac{0.0012}{e_2}

          0.5 e_2 =  0.0012

         e_2 = \frac{ 0.0012  }{0.5}

          e_2 = 0.0024 \  m =  2.4 mm

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