Question

A bullet is fired into the air with an initial velocity of 1,200 feet per second at an angle for 45° from the horizontal. What is the horizontal distance traveled by the bullet in 2 seconds? (Neglect the resistance of air on the bullet. Round your answer to the nearest whole number.)
___ ft

Answer 1

Hint:
Ignoring air resistance:
1. Horizontal component velocity = initial velocity* cos(theta)
where theta is angle with horizontal.
2. Horizontal velocity is constant and equal to velocity calculated in 1. above.

Answer 2

@mathmate I got 630 into that equation rounded, but that answer is wrong..

Answer 3

@zepdrix

Answer 4

630 is not the correct answer.
It is not the horizontal velocity either.
The question requires the horizontal \(distance\) travelled after 2 seconds.

Hints:

Horizontal velocity, Vh
= initial velocity * cos (theta)
=1200*cos(45) ft/s

Distance = velocity (speed) * time
= Vh* time

where time = 2 seconds.
Answer should be in feet.

Answer 5

I'm still not understanding.. @mathmate

Answer 6

Horizontal velocity of a projectile is not subject to acceleration due to gravity, so the projectile travels horizontally with a uniform velocity, Vh.
where
Vh=Vi cos(theta) (see diagram),
where
Vi=initial velocity=1200 ft/s.

After that, use distance = Time * speed/velocity to get the horizontal distance.

For more detailed notes, read your teacher's notes, or try:
http://www.physicsclassroom.com/class/vectors/Lesson-2/Non-Horizontally-Launched-Projectiles-Problem-Solv

Answer 7

So 1200 cos (45)
then 2 x that number? @mathmate

Answer 8

Yes, that is correct!
BUT, you need to understand why this is done, or else you will not be able to solve similar problems.
Read the link I provided above, and if you need more explanations, post your question.