A glaucous-winged gull, ascending staight upward at 5.20 m/s,drops a shell when

Question

A glaucous-winged gull, ascending staight upward at 5.20 m/s,drops a shell when it is 12.5 m above the ground. A) What is the magnitude and direction of the shell'sacceleration just after it is released? B)Find the maximum height above the ground reached by theshell. C)How long does it take for the shell to reach theground? A glaucous-winged gull, ascending staight upward at 5.20 m/s,drops a shell when it is 12.5 m above the ground. A) What is the magnitude and direction of the shell'sacceleration just after it is released? B)Find the maximum height above the ground reached by theshell. C)How long does it take for the shell to reach theground?

Explanation / Answer

Upward speed v = 5.2 m / s (a).  the magnitude of the shell's accelerationjust after it is released a = -9.8 m / s ^ 2 And it is act downward direction B)The maximum height reached by the shell above 12.5m is H = v ^ 2 / 2g                                                                                                      = ( 5.2 ) ^ 2 / ( 2 * 9.8 )                                                                                                      = 1.379 m So, the maximum height above the ground reached by the shell H' = H + 12.5                                                                                                    = 13.879 m (C). time taken for the shell to reach the groundT = t + t ' + t " where t = time taken to reach 12.5 m             = 12.5 m / 5.2 m / s              =2.403 s           t ' =time taken to reach H ' height from height H              =v / g             = ( 5.2 m / s ) / ( 9.8 m / s ^ 2 )             = 0.5306 s          t " = timetaken to reach the shell from H ' to ground             = [ 2 H ' / g ]             = 2.832 s So, T = 5.766 s

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