Finding the angle of the projectile

[nagaromo]

Homework Statement

A gun has a muzzle velocity of 300m/s. If you want to hit a target that has a horizontal distance of 1.00km away and is 150m above you

a)What angle does the line between you and the target make with respect to the horizontal?
b)At what angle should you aim?

Homework Equations

x=vcos$$\theta$$t and y= -1/2at²

The Attempt at a Solution

a)For this one I used arc tan and I got 8.53.
b)From here my head wanted to explode because I have no idea how to do it. I tried to make up my own angles. I tried using x=vcos$$\theta$$t and y= -1/2at² , but I realized it wouldn't work because I have two variables. Thank you very much for helping/attempting to help![/b]

 

[Twoism]

Two big hints I will give you:
You want your max height of the bullet, in the y direction, to be 150m, where the target is.
Assuming 150m is your max height, what does that tell you about the velocity of the bullet, more specifically Viy and Vfy?
With that in mind, what does the equation Vf^2=Vi^2 +2ad tell you in the y direction and how does that help you?

 

[nagaromo]

umm i set vf^2 = 0 so that it'll reach 150 as the max height and solved for theta and i got 10.37 as the angle. idk if that's correct, but now i have a starting point. thank you so much! :D

 

[Twoism]

umm i set vf^2 = 0 so that it'll reach 150 as the max height and solved for theta and i got 10.37 as the angle. idk if that's correct, but now i have a starting point. thank you so much! :D

10.37 degrees is the right answer.
You're welcome.
EDIT: I would also note that theta could actually be another value too, but the one you got is the most efficient and most direct.

 

[zgozvrm]

10.37 degrees is the right answer.

Really?

If the velocity is [itex]V_i = 300 m/s[/tex] and the angle of trajectory is [itex]\theta = 10.37^\circ[/tex] then that makes<br /> <br /> $$V_{ix} = V_i cos(\theta) = 295.1 m/s$$<br /> <br /> and<br /> <br /> $$V_{iy} = V_i sin(\theta) = 54.0 m/s$$<br /> <br /> <br /> Therefore the time necessary to hit the target would be found by <br /> <br /> $$t = \frac{Dx}{V_{ix}} = \frac{1000}{295.1} = 3.39 sec$$<br /> <br /> <br /> The vertical distance would then be found by<br /> <br /> $$D_y = V_{iy}t + \frac{1}{2}at^2 = (54)(3.39) + (0.5)(-9.8)(3.39)^2 = 126.7 m$$<br /> <br /> <br /> Sounds to me like you aimed a bit low...[/itex][/itex]

 

[zgozvrm]

You want your max height of the bullet, in the y direction, to be 150m, where the target is.

You shouldn't assume that the bullet has reached it's maximum possible height at that trajectory angle. The reason the bullet doesn't fly any higher is because it was stopped by the target.

Think about throwing a ball at a wall 3 meters away at an angle of, let's say, 40[itex]^\circ[/tex] and a velocity of 25 m/s.<br /> Does the fact that the ball hit the wall at a specific height mean that the ball reached it's maximum height?<br /> <br /> No. The wall stopped it from reaching it's maximum height. The ball still had vertical velocity - In fact, the ball would hit the wall in 0.16 sec. Had the wall not been there, it would have taken 3.28 sec for the ball to reach it's peak vertical height. <i>That's</i> when the vertical velocity is 0 m/s.[/itex]

 

[nagaromo]

You shouldn't assume that the bullet has reached it's maximum possible height at that trajectory angle. The reason the bullet doesn't fly any higher is because it was stopped by the target.

Think about throwing a ball at a wall 3 meters away at an angle of, let's say, 40[itex]^\circ[/tex] and a velocity of 25 m/s.<br /> Does the fact that the ball hit the wall at a specific height mean that the ball reached it's maximum height?<br /> <br /> No. The wall stopped it from reaching it's maximum height. The ball still had vertical velocity - In fact, the ball would hit the wall in 0.16 sec. Had the wall not been there, it would have taken 3.28 sec for the ball to reach it's peak vertical height. <i>That's</i> when the vertical velocity is 0 m/s.[/itex]

$<br /> How would you solve this question, zgozvrm? I'm totally confused.$

 

[zgozvrm]

Well, you already know:

the horizontal distance [itex]D_x[/tex] = 1000m<br /> the vertical distance [itex]D_y[/tex] = 150m<br /> the initial velocity [itex]V_i[/tex] = 300 m/s<br /> the acceleration of gravity g = 9.8 m/s/s<br /> <br /> <br /> <br /> Given the 4 basic kinematic equations only one doesn't involve the final velocity [itex]V_f[/tex]:<br /> $$D = V_it+at^2$$<br /> so let's use that...<br /> <br /> <br /> $$D_y = V_{iy} t + a t^2$$<br /> $$D_x = V_{ix} t + a t^2$$<br /> <br /> We know there's no acceleration in the horizontal direction, so that leaves us with<br /> $$D_x = V_{ix}t$$<br /> <br /> We don't know the time, so solve that equation for t then substitute into the equation for [itex]D_y[/tex]<br /> <br /> The resulting equation will have the following variables: [itex]D_y[/tex], [itex]D_x[/tex], [itex]V_{iy}[/tex], [itex]V_{ix}[/tex], and [itex]a[/tex]<br /> The only variables we don't know are the initial vertical velocity [itex]V_{iy}[/tex] and the initial horizontal velocity [itex]V_{ix}[/tex].<br /> <br /> But we know that [itex]V^2 = V_{ix}^2 + V_{iy}^2[/tex], so solve the previous equation for [itex]V_{iy}[/tex] and substitute into this equation.<br /> <br /> Solve the resulting quadratic for [itex]V_{ix}[/tex]<br /> <br /> From there, you should be able to determine [itex]V_{iy}[/tex] and the angle of trajectory [itex]\theta[/tex][/itex][/itex][/itex][/itex][/itex][/itex][/itex][/itex][/itex][/itex][/itex][/itex][/itex][/itex][/itex][/itex][/itex]