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A body is projected at time (t = 0) from a certain point on a horizontal planet surface with a certain velocity at a certain angle with the planet`s surface (assumed horizontal). The horizontal and vertical displacement x and y in metre are related to time as x = 10root3t and y = 10t - 4t^2. find the vertical component of velocity of the particle when it is at a height half of the maximum height attained.

A body is projected at time (t = 0) from a certain point on a horizontal planet surface with a certain velocity at a certain angle with the planet`s surface (assumed horizontal). The horizontal and vertical displacement x and y in metre are related to time as x = 10root3t and y = 10t - 4t^2. find the vertical component of velocity of the particle when it is at a height half of the maximum height attained.

Grade:12th pass

1 Answers

Vikas TU
14149 Points
7 years ago
Since vertical height eqn. is given which is:
y = 10t - 4t^2
we find when it attains max. height.
Therefore, dy/dt = 10 – 8t = 0
t = 10/8 = > 5/4 = > 1.25 seconds.
put t in y eqn. we get,
ymax. = 10*1.25 – 4*1.25*1.25
   = 12.5 – 6.25
   = 6.25 m.
 
at half of 6.25 that is => 3.125 m
solving with 3.125 m in y = 10t - 4t^2.
3(approx.) = 10t - 4t^2
t = 0.38 seconds.
 
Hence at t= 0.38 velocity would be=>v = dy/dt = 10 – 8t = > 10 – 8*0.38 = > 10 – 3.8 = > 6.2 m/s (approx.)

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