Launch angle?

[Gareth56]

This one has me puzzled:-

 

At 1/2 of its maximum height, the speed of a projectile is 3/4 of its initial speed. What was its launch angle?

[timo]

What's your ideas? Have you tried approaching the problem? How did you do that? What was the problems you encountered?

[MrMongoose]

Horizontal speed is equal to the secant of the angle of elevation of the velocity. The rest is simple equations of motion, available at any good amths teacher.

[JaKiri]

The best way to do most mechanics problems at this level is to sketch a diagram, (ideally without the huge amount of dead space I have on mine) eg:

 

 

you know that acceleration in the, horizontal, x direction, x'' is zero. Acceleration in the verticle, y, direction, y'' is -g.

 

You can work out the initial velocities from basic trigonometry, the initial positions are at 0 and 0, and you now have all the information you need to plug into the kinematics equations to get equations for how the horizontal and vertical distances, x and y, and velocities, x' and y', change with respect to time.

 

When the speed of the projectile is 3/4 its original, the squareroot of x'² + y'² will be equal to 3v/4 (on my diagram). You can find out how much half the height is using the kinematics equations, equate them and you get an equation in phi, which gives you the answer.

[Gareth56]

Thanks for the pointers JaKiri.

 

I've written down all the equations that should be relavent ie

 

Ux=U*Cos alpha

Uy=U*Sin alpha

 

 

then y = Uyt -1/2gt^2

x = Ux * t

 

Vy = Uy -gt

Vx = Ux

 

 

That's as far as I get but it doesn't really matter it was just a querry regarding a question I came across in Tipler regarding projectile motion. After the above I think it's all algebra or whatever. In case certain people are wondering it is not any kind of assignment question, it's just that I have an interest in physics...I'm well past college days

[JaKiri]

Looking back, it's a bit of a pathetic diagram all round. I should really have drawn the acceleration due to gravity on it.

[Norman Albers]

I wrote four lines of energy equation, identifying vertical velocity which changes, and horizontal velocity which does not. The clue about being halfway up can be expressed simply in terms of the square of vertical velocity, yah?

[Severian]

The easiest way to do it is to think about energy.

 

At launch the potential energy V=0, while the kinetic energy is [math]K = \frac{1}{2}mv^2=\frac{1}{2}m(v_x^2+v_y^2)[/math] (where [math]v_{x,y}[/math] are the horizontal and vertical components of the initial velocity.)

At maximum height h, V=mgh and [math]K=\frac{1}{2}mv_x^2[/math]

At half maximum height V=mgh/2 and [math]K=\frac{1}{2}m\left(\frac{3}{4}v\right)^2=\frac{9}{32}mv^2[/math]

 

Conservation of energy tells us the sum of the two is constant, so

 

[math]\frac{1}{2}m(v_x^2+v_y^2) = mgh + \frac{1}{2}mv_x^2 = \frac{1}{2}mgh + \frac{9}{32}m(v_x^2+v_y^{2})[/math]

 

This has 2 equalities. Use the first to determine h:

[math]mgh=\frac{1}{2}mv_y^2[/math]

Stick this in the second:

[math]\frac{1}{2}v_y^2 + \frac{1}{2}v_x^2 = \frac{1}{4}v_y^2 + \frac{9}{32}(v_x^2+v_y^{2}) \Rightarrow \frac{7}{32}v_x^2 = \frac{1}{32}v_y^{2}\Rightarrow \frac{v_y}{v_x} = \sqrt{7}[/math] and the angle is [math] \tan^{-1} \sqrt{7}[/math]

[Norman Albers]

Yes! At any point "speed" means total velocity which is the vector sum of the two components. Horizontal velocity is assumed to not change, and vertical velocity is related through its square to energy and height.

[MrMongoose]

Yes! At any point "speed" means

 

[magnitude of]

 

total velocity which is the vector sum of the two components. Horizontal velocity is assumed to not change, and vertical velocity is related through its square to energy and height.

 

Just to be clear