Projectile Motion and Impact: Solving for Elevation Angle and Time Interval

[Jacobpm64]

Homework Statement

A gun on the shore (at sea level) fires a shot at a ship which is heading directly toward the gun at a speed of 40 km/h. At the instant of firing, the distance to the ship is 15,000 m. The muzzle velocity of the shot is 700 m/s. Pretend that there is no air resistance.
(a) What is the required elevation angle for the gun? Assume g = 9.80 m/s^2.
(b) What is the time interval between firing and impact?

Homework Equations

Uhmm, I'm guessing. t_flight = \frac{2v_{0} sin\alpha }{g}

Maybe x_max = \frac{v^{2}_{0}sin2\alpha}{g} will also be applicable.

v_{x} = v_{0x} = v_{0}cos\alpha
v_{z} = v_{0z} - gt = v_{0} sin \alpha - gt
x = v_{0x}t
z = v_{0z}t - \frac{1}{2} gt^2

The Attempt at a Solution

I'm not sure what to do with this problem. Any way I set it up, I end up with more variables than I can solve for.

I would appreciate a hint to throw me in the right direction. Thanks in advance.

 

[cristo]

You seem to be throwing in equation willy-nilly. Draw a diagram, and do the usual thing of splitting up into horizontal and vertical components, and use the kinematic equations. Your horizontal target will be moving, so x_max=1500-40t (original distance minus the distance the ship covers in the time the bullet is in the air). You should be able to write an equation for the x component of the displacement of the bullet. Eliminate t using an equation for the y component of the displacement of the bullet.

 

[Jacobpm64]

40 \frac{km}{h} * 1000 \frac{m}{km} * \frac{1}{3600} \frac{h}{s} = 11.1 \frac{m}{s}

x_{max} = 15,000 - 11.1 t

x(t) = 700 cos \alpha t

700 cos \alpha t = 15000 - 11.1 t

z(t) = \frac{-1}{2} (9.80) t^2 + 700sin \alpha t

I don't know what to do now.. But I'm pretty sure my equations are good.

 

[cristo]

Ok, well use the fact that the displacement, z, is equal to zero when the bullet hits the ship. Can you then solve this equation for t? (Hint: quadratic formula)

 

[Jacobpm64]

0 = \frac{-1}{2} (9.80) t^2 + 700 sin \alpha t

By the quadratic equation,

t = 142.86 sin \alpha

So plugging t in.. I get..
100000 sin \alpha cos \alpha = 15000 - 1587.302 sin \alpha

How do I solve this for \alpha ?

 

[jfrenchie]

Use the 2 graph method, I just got this problem for homework myself. The angle should come out to around 8.6, and the time to around 21.4 seconds. Hope this helps