SOLUTION: You fire a rifle at an angle of 45 degrees. Thus the initial horizontal and vertical velocities of your bullet are the same. Suppose they each equal 240 feet per second. Again igno

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Question 280279: You fire a rifle at an angle of 45 degrees. Thus the initial horizontal and vertical velocities of your bullet are the same. Suppose they each equal 240 feet per second. Again ignore air resistance. Assume you are shooting from ground level (height 0). Your bullet will hit the ground ____ feet from your current position.
Found 2 solutions by nerdybill, jsmallt9:
Answer by nerdybill(7384) (Show Source):
You can put this solution on YOUR website!
You fire a rifle at an angle of 45 degrees. Thus the initial horizontal and vertical velocities of your bullet are the same. Suppose they each equal 240 feet per second. Again ignore air resistance. Assume you are shooting from ground level (height 0). Your bullet will hit the ground ____ feet from your current position.
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This is really a physics problem.
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Applying the equation of motion:
y = (1/2)(-g)t^2 + vt + yi
where
g is gravity 32 feet per second squareed
t is time in secs
v is initial velocity
yi is the initial height (0)
.
0 = (1/2)(-32)t^2 + 240t + 0
0 = -16t^2 + 240t
0 = -t^2 + 15t
0 = t^2 - 15t
0 = t(t - 15)
t = 15 seconds (time it will hit ground)
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equation of motion in the x direction:
x = (1/2)at^2 + vt + xi
Since acceleration in the x direction is zero
and, initial start is zero we have
x = vt
x = 240(15)
x = 3600 feet

Answer by jsmallt9(3758) (Show Source):
You can put this solution on YOUR website!
The problem does not mention friction. So I assume that we are allowed to disregard it.

Without friction the distance from the current position where the bullet will hit the ground will simply be horizontal speed, 240, times the number of seconds the bullet is in the air. So we need to find the number of seconds the bullet is in the air.

The time the bullet is in the air is determined by how long it takes gravity to pull the bullet bask to the ground. For this we will use the general equation

where
a = acceleration. The only acceleration in this problem is the acceleration due to gravity. If distance above ground is considered positive, then the gravity will be a negative because it is pulling down (the negative direction). Since the problem uses feet we will use -32 ft/sec/sec for "a".

t = time (in seconds)
= the initial velocity. The initial vertical velocity is 240 ft/sec.
= the initial position (height). We are told to use zero for this.

So the equation for vertical motion in this problem is:

which simplifies to:

We can use this equation to find the time(s) when s(t) is zero (i.e. the bullet is at ground level):

This is a quadratic equation so we will factor it:

and use the Zero Product Property:
-16t = 0 or t - 15 = 0
t = 0 or t = 15
This tells us that at time t = 0 the bullet was at ground level (which we already knew) and that at time t = 15 the bullet will again be at ground level.

Now we know that the bullet will be in the air for 15 seconds. This means the bullet will be 240(15) or 3600 feet away.