A hot-air balloonist, rising vertically with a constant velocity of magnitude 5.00 m/s, releases a sandbag at an instant when the balloon is 40.0 m above the ground. After the sandbag is released, it is in free fall. Find the position of the sandbag at 0.250 s after its release.

Respuesta :

Answer:

y=39.057 m

Explanation:

Using Kinematic relation

[tex]s=ut+ \frac{1}{2}at^2[/tex]

given u= 5m/s

a=g= -9.81 [tex]m/s^2[/tex]( directed downward)

[tex]s=5t- \frac{1}{2}(9.80)t^2[/tex]

Also, we know that

v=u+at

v=5-9.80t

at time t= 0.250 sec

[tex]s=5\times0.25- \frac{1}{2}(9.80)0.25^2[/tex]

s=0.94375 m

now position of sandbag

y= 40-0.94375

y=39.057 m

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