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A particle is projected vertically upwards from a point A on the ground. It takes time t1, to reach a point B, but it still continues to move up. It takes further time t2 to reach the ground from point B then height of point B from the ground is: (a) [latex]frac{1}{2}g(t_{1}+t_{2})^{2}[/latex] (b) [latex]gt_{1}t_{2}[/latex] (c) [latex]frac{1}{8}g(t_{1}+t_{2})^{2}[/latex] (d) [latex]frac{1}{2}gt_{1}t_{2}[/latex]

Given Total time of particle = (t1 + t2) To find initial value Applying 2nd equation of motion We get, Since height at t = t1+ t2 = 0 At t =... View Article

A particle moving along x-axis has an acceleration f at time t, given by [latex]f = f_{0}(1-frac{t}{T})[/latex], where [latex]f_{0}[/latex] and T are constants. The particle at T = 0 has zero velocity. In the time interval between t = 0 and the instant when f = 0, the velocity (va) of the particle is then:(a) [latex]frac{1}{2}f_{0}T[/latex](b) [latex]f_{0}T[/latex](c) [latex]frac{1}{2}f_{0}T^{2}[/latex](d) [latex]f_{0}T^{2}[/latex]

Given and T are constants At t = 0 and v = 0 When f = 0 then, 0 = T = t ------------ (1) f = (final - initial velocity)/ Time From equation... View Article