Help with a very easy mechanic problem. (Still I can´t solve it)

[mprm86]

How should Vo and Theta should be chosen (and why), in order to making the distance X minimum? Consider that theta can be any number between 0 and 90 and Vo can go from 0 to a maximum value.

 

[tnorkhangpa]

See, x will be minimum if the projectile(assuming it is a point- object) just crosses the edge of the table. Just try to equate the range,(R)
R = 2v^2(sin(theta)cos(theta)/ g
You know the value of R(which is given in the question). Just use that and you will know the value of theta and initial velocity.

 

[mprm86]

The problem is for any initial velocity, not for an specific one. It is also for any value of Theta. The problem here is to see how should we choose theta and Vo, for making the distance x as minimum as possible.
I guess that if we chooose that Vo be the maximum it can be, so the height it reaches will be also maximum for values of theta near to 90. The problem here is, which should be the exact value of theta, when Vo is maximum, in order to making the distance x the minimum.

 

[dextercioby]

Okay then,so you've got the function R(v_{0},\theta) which needs to be minimized.Okay,then treat is as a maths problem in which you've got to find out the values for extremum for a 2-variable function.

Set the PD wrt the variables equal to 0,then find the values for extremum and then compute the hessian on these solutions.It should be >0,for a minimum.

Daniel.

 

[Andrew Mason]

How should Vo and Theta should be chosen (and why), in order to making the distance X minimum? Consider that theta can be any number between 0 and 90 and Vo can go from 0 to a maximum value.

The trajectory is a parabola. So the drawing is not accurate.

Express y as a function of x (hint: express t in terms of x and substitute into the expression for y as a function of t).

The minimum X distance will result when the projectile just misses the table. So there are two points: the take-off point and the end of the table. If v0 is not variable, then there is only one parabola that fits those two points. If v0 is variable, you want the parabola that has the greatest slope dy/dx as it passes the end of the table. That would be the parabola that has the greatest vertical speed at that point - ie the one that reaches the greatest height.

AM

 

[mechanicsbook]

X can be 0 if teta is90 and Vo is infinite so I consider the problem is not very well established,

 

[cincirob]

It seems to me the problem intends to ask what velocity produces the minimum x for a given angle between 0 and 90 degrees. It wil become clear why as the solution is given. The minimum x will occur for a given angle when the trajectory is such that the projectile just misses the corner of the step. The time it takes to traverse the hoizontal distance R is T(h) = R/(Vo*cos (theta)). The vertical component of velocity at that point in time will be the negative of the initial vertical velocity, so V=Vo + aT
-Vo=Vo + aT >> Vo= -aT/2. Substituing yields
Vo = -(-32.2ft/s/s)(R/(Vo*cos (theta))/2
Vo = (32.2*cos(theta)/2R)^.5
If I did the algebra correctly...be sure to check it, this will give Vo for a given theta. Since the cosine of 90 degrees is infinite, Vo could be infinite. That's the reason I chose to work it as a function of theta.

 

[Andrew Mason]

It seems to me the problem intends to ask what velocity produces the minimum x for a given angle between 0 and 90 degrees. It wil become clear why as the solution is given. The minimum x will occur for a given angle when the trajectory is such that the projectile just misses the corner of the step. The time it takes to traverse the hoizontal distance R is T(h) = R/(Vo*cos (theta)). The vertical component of velocity at that point in time will be the negative of the initial vertical velocity, so V=Vo + aT
-Vo=Vo + aT >> Vo= -aT/2. Substituing yields
Vo = -(-32.2ft/s/s)(R/(Vo*cos (theta))/2
Vo = (32.2*cos(theta)/2R)^.5

Should be:
v_0 = -aT/2 = -gR/2v_0cos(\theta)

v_0 = -\sqrt{gR/2cos\theta}

So for a given initial speed, x will be minimum when:

cos\theta = -gR/2v_0^2


AM