The horizontal and vertical distances travelled by a particle in time t are given by x =6t and y=8t-5t^2
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I think your question is ------> The height y and the distance x along horizontal, for a body projected in the vertical plane are given by; y=8t-5t2 and x=6t . What is its initial velocity? Find by calculus method?
solution :- x = 6t
differentiate x with respect to time t
dx/dt = 6 = x - component of initial velocity =Vx
y = 8t - 5t²
differentiate y with respect to time t
dy/dt = 8 - 10t
At t = 0, dy/dt = 8 = y - component of intial velocity = By
Now, |V| = √(Vx² + Vy²) = √(6² + 8²) = 10m/s
solution :- x = 6t
differentiate x with respect to time t
dx/dt = 6 = x - component of initial velocity =Vx
y = 8t - 5t²
differentiate y with respect to time t
dy/dt = 8 - 10t
At t = 0, dy/dt = 8 = y - component of intial velocity = By
Now, |V| = √(Vx² + Vy²) = √(6² + 8²) = 10m/s
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