A projectile is fired horizontally from a gun that is 45.0 m above flat ground, emerging from the gun with a speed of 250 m/s. (a) How long does the projectile remain in the air? (b) At what horizontal distance from the firing point does it strike the ground? (c) What is the magnitude of the vertical component of its velocity as it strikes the ground?

Question

phi · Answer

Do you know how far an object falls due to gravity?

anonymous · Answer

9.8m/s^2

phi · Answer

That is acceleration.  you need the formula for the distance an object falls when its acceleration is 9.8 m/s^2

anonymous · Answer

is it y-y0= voy(-1/2gt^2)

anonymous · Answer

?

anonymous · Answer

i used it but get the wrong answer

phi · Answer

distance = 1/2 g t^2     is how I think of it.  The object starts at 45 m high and falls. so
solve
45 = 1/2 * 9.8 m/s^2 * t^2

anonymous · Answer

oh so you don't include the 250m/s ahhhhh

anonymous · Answer

i guess that is x velocity

anonymous · Answer

so thats why...bah

phi · Answer

If you shot down (or up) , then you have to account for that,  but horizontal the object goes sideways and falls at the same time.

anonymous · Answer

excellent. thanks

anonymous · Answer

ok do you know how to approach c?

phi · Answer

V= a*t  or in this case V= gt

phi · Answer

For (a) you should get something very close to 3 sec  (3 if g=10 exactly)
(b) close to 750 m
(c) close to 30 m/s