# More projectile motion problems (sigh)

1. Mar 13, 2006

### raizen91

got two more problems, hope you could help. more projectile problems...(sigh), i dunno how to start, hoping you guys could help me with the initial steps...

1. a missile is fired with a launch velocity of 15,000 ft/s at a target 1,200 miles away. at what angle must it be fired to hit the target? how long after it is fired will the target be hit.

2. a projectile is fired at an angle of 30 degrees above the horizontal from the top of a cliff 600 ft. high : the initial speed of the projectile is 2000 ft/s. how far will the projectile move horizontally before it hits level grained at the base of the cliff?

2. Mar 14, 2006

### plusaf

1) no air resistance
2) gravity is that of earth
3) can you write an equation that describes the motion of the projectile that can be solved for the unknown you're looking for?

um... and the first one is worded poorly. "missiles" are usually "fired" or launched, but then they continue to burn their own fuel and propel themselves toward their targets :) thus, for question 1), the answer could well be "zero degrees" of launch angle, with the assumption that the thing's got wings and a guidance control system. )) but not likely.
better to assume it's "fired" like a projectile, much as in problem #2, with an initial velocity with horizontal and vertical components, and go from there.

3. Mar 14, 2006

### topsquark

For 1 you might find this thread useful: https://www.physicsforums.com/showthread.php?t=112308

-Dan

4. Mar 15, 2006

### raizen91

ok, here's what i did.
vx = r / t
dx/ v[subx] = vx * t / vx
= 1931,216.66 m / 4572.009
= 422.4 seconds
now for the angle
cos[theta] = vx / vx
= 4572.009 / 4572.009 =1
= 60 degrees
ok...is this right?

5. Mar 16, 2006

### Staff: Mentor

Not clear to me what you are doing here. Start by expressing the horizontal and vertical components of the initial velocity in terms of the initial speed (which is given) and the unknown angle. (Do this algebraically; don't plug in numbers until the last step.) Then:

(1) Write an expression for horizontal distance as a function of time. (Use $x = x_0 + v_0 t[/tex].) (2) Write an expression relating the initial vertical speed with the total time of flight. (Use [itex]v = v_0 + a t$.)

You'll get two equations and two unknowns; you'll be able to solve for the angle and the time. When you finally plug in numbers, be sure to use proper units. (Convert distances to feet.)

6. Mar 17, 2006

### michealsmith

silly wording but ithink its clear...i doubt they r refering to some sorta ..projectile mathematics...use eqautions of motion....sumbody do it....my hands r lazy..ok ..good luck..

7. Mar 17, 2006

### hotvette

Click the Basic Motion tutorial in my signature for some help with methodology.

8. Mar 17, 2006

### Da-Force

Let's walkthrough real quickly.

1. a missile is fired with a launch velocity of 15,000 ft/s at a target 1,200 miles away. at what angle must it be fired to hit the target? how long after it is fired will the target be hit.

The velocity must be split up into two components. X and Y. Cos@ and Sin@ respectively.

Vix=15000cos@
X=1200 miles (6336000 feet)
Since there is no acceleration horizontally, the equation is X=Vix*t.

Viy=15000sin@
Y=maximum height which occurs halfway (600 miles) at t/2.
Since there is acceleration vertically downward, the equation is:
Y=Viy*t+1/2*g(t^2)

Note that this equation for the Y direction is a parabolic equation and has two zeros. Initial point and the point where it hits.

2. a projectile is fired at an angle of 30 degrees above the horizontal from the top of a cliff 600 ft. high : the initial speed of the projectile is 2000 ft/s. how far will the projectile move horizontally before it hits level grained at the base of the cliff?

This is an easier problem than number one.

@=30 degrees.
Vi=2000 ft/s
Vix=2000cos30
Viy=2000sin30
X=?
Y=600 ft

Y=Viy*t + 1/2*g*t^2
X=Vix*t

Find the total time it takes for it to hit the ground (hint, use the equation for the Y direction and then plug in t for the equation for X direction)

9. Mar 18, 2006

### PhY_InTelLecT

For qn 1, what you could actually do was to solve the question using the eqns for both the horizontal and vertical components. Assume no air resistance, ur horizontal component eqn will be simply s=ut while the eqn for vertical component will be s=ut+1/2at^2.. Find t in terms of s and u using the the first eqn and sub the t into ur 2nd eqn..and thus, simply solve for
the unknown angle..

Pls note that that ur vertical speed must also be expressed in the (sin) or (cos) of the unknown angle.

Last edited: Mar 18, 2006