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A ball is dropped off the side of a bridge.
After 1.55 s, how far has it fallen?
(Unit = m)
Remember: Falling = -V
Lost height = -Ay
a = -9.80 m/s2 Be careful with minus signs!!!

Respuesta :

ashish4112119

second equation of motion

s = ut + 1/2at^2

s is the distance travelled time t

u is the initial velocity

t is the time

a is the acceleration of the body in motion

u is 0

s = 1/2 * -9.8 * (1.55)^2

s = -11.77 m

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