During the town fiesta, Vingie who is on top of the Palo Sebo drops the prize bag to be caught by his friends below. If the bag falls to the ground after 3.00 secs, how high up was the boy?

Question

anonymous · Answer

Given:
g= -9.8 m/s^2
Vi= 0 m/s
dy= ?
t= 3.0 s

use the formula:
distance= Vi(t) + 1/2g(t)^2

substitute the given values:
distance= (0m/s)(3.0 s) + 1/2(-9.8 m/s^2)(3.0 s)^2
               = (0 m) + 1/2(-9.8 m/s^2)(9.0 s^2)
               = (0m) + 1/2(-88.2 m)
               = 0m+ (-44.1 m)
               = 44.1 m
*the negative sign indicates the direction on which the bag falls. since there isn't any negative distance, the negative sign is omitted.

THEREFORE, THE BOY IS 44.1 m ABOVE THE GROUND.

anonymous · Answer

LOL :))))
vingie? gievin TAHAHAHAHAHAHA =))))))))

anonymous · Answer

LOL YEAH

anonymous · Answer

h=.5 g t^2
=.5*9.8*9=44.1 m