A particle moves along parabolic path y=2x-x²+2 in such a way that the x component of velocity vector remains constant(5m/s). Find the magnitude of acceleration of the particle.

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Solution

y=2x-x²+2
Particle moving in a parabola under force of gravity follow the below given equation.
y = x tan Ф - g x²/(2 u² Cos² Ф) + yo
yo = Initial y co-ordinate

Comparing the two equations,
tanФ = 2
and g /(2 u² Cos² Ф) = 1
It is given that horizontal component of velocity = 5m/s
=> ucos(Ф) = 5m/s

and g /(2 u² Cos² Ф) = 1
=>g = 2 u² Cos² Ф
= 2(ucosФ)²
= 25*5
= 50

Acceleration in y- direction = 50

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A particle moves along parabolic path y=2x-x2 ​​​​​+2 in such a way that the x component of velocity vector remains constant(5m/s). Find the magnitude of acceleration of the particle.

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