After a ball rolls off the edge of a horizontal table at time t = 0, its velocit

Question

After a ball rolls off the edge of a horizontal table at time t = 0, its velocity as a function of time is given by v rightarrow = 1.1 i^ - 9.8t j^ where v rightarrow is in meters per second and t is in seconds. The ball's displacement away from the edge of the table, during the time interval of 0.495 s for which the ball is in flight, is given by delta r rightarrow = v rightarrow dt. To perform the integral, you can use the calculus theorem A + B f(x)] d x = A dx + B f(x) dx. You can think of the units and unit vectors as constants, represented by A and B. Perform the integration to calculate the displacement of the ball from the edge of the table at 0.495 s. delta r rightarrow = m

Explanation / Answer

r=1.1(.495)i+(-9.8)(t^2/2).4950j

=.5445i-1.2j m

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