Question

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?

Answer 1

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.

Answer 2

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

Answer 3

LOL YEAH

Answer 4

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