a 55.0 g bullet is fired vertically with an initial velocity of 123m/s

1) what is the velocity at it's highest point of travel?
2) what maximum height does it reach?
3) at what velocity does it hit the ground?

Respuesta :

1) at highest point of travel it stops. and there v=0 m/s
3) when it returns to the original point it's speed would be again 123m/s
(conservation law and no calculations needed at all)

2) this one is harder

step 1 - finding the time taken from ground to highest point 
v = u + at
v is the speed at highest point (it's 0, as it stops) 
u - speed at origin (123m/s given)
a - acceleration (in our case it's gravitation and it quals 9.8m/s, some people use just 10 m/s, by the way, it's negative)
0 = 123 + (-9.8)* t
t = 123/9.8 = say 12.3 second
it's time taken to travel up

step 2 - finding displacement or heigh
original formula is 
v^2 = u^2 + 2ar
v - speed at point (0 m/s at the top point)
u - initial speed (123m/s)
a - acceleration (it's gravitation = 9.8 or just 10) and it's again negative
r - displacement or height or distance travelled 
0^2 = 123^2 + 2*(-10)*r
123*123=20r
r= 123*123/20
r= 756 meters 

there is also mass given. it's just to confuse you. no matter that the mass is - time taken, max height or range, acceleration and velocity would be the same for any mass. surprised?