Tardigrade - CET NEET JEE Exam App
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Tags: Class 11 , Physics , Laws of Motion Asked by Vagmita Pandya
Open in App Suggest Corrections Text Solution Solution : To find the force, we need to find the acceleration experienced by the particle. <br> The acceleration is given by `a = (d^2 y)/(dt^2) " or " a = (dv)/(dt)` <br> Here v= velocity of the particle in y direction ` v = (dy)/(dt) = u - "gt"` <br> The momentum of the particle ` = mv = m ( u - "gt" )` <br> `a = (dv)/(dt) = - g` <br> The force acting on the object is given by F = ma = -mg <br> The negative sign implies that the force is acting on the negative y direction. This is exactly the force that acts on the object in projectile motion. Text Solution Solution : Here, ` y = ut + (1)/(2) g t^(2) ` <br> ` velocity , upsilon = (dy)/(dt)=u+ (1)/(2) g (2t)= u + g t , ` <br> acceleration `a = (du)/(dt) = 0 + g = g ` <br> ` As F= ma :. F = mg ` <br> Thus the given equation describes the motion of a partical under acceleration due to gravity , and y is the position co- ordinate in the direction of g . |