Projectile Motion with initial velocity

[Yashbhatt]

Homework Statement

A projectile with initial velocity u making an angle A with the horizontal. Find the time t when the velocity of the projectile becomes perpendicular to its original velocity.

Homework Equations

v = u + at
s = ut + 1/2at^2
v^2 - u^2 = 2as[/B]

The Attempt at a Solution

I have an intuitive idea of how to solve it. When the velocity becomes perpendicular to the original velocity, their dot product will be zero. I don't know if this is the right way. If this the right way, I don't know how to solve for it.

 

[Orodruin]

That indeed sounds like a good approach. So how can you realize your idea? What is the velocity vector at a given time?

 

[Yashbhatt]

It's the vector sum of v(y) = u sinA - gt and v(x) = u cosA

 

[Orodruin]

Indeed. So what is the inner product of that vector at $t=0$ with that vector at $t=T$?

 

[Yashbhatt]

That's were the problem is. I got it as the square root of v^2 - 2 u g t sinA + g^2 t^2. I don't know what to do further.

 

[Orodruin]

That looks as the square root of the actual inner product (assuming the v^2 should be u^2), but it should be zero anyway so this does not matter much.

So, do you know how to solve for the roots of a second order polynomial?

 

[Yashbhatt]

Through the Discriminant?

 

[Orodruin]

The discriminant gives you some information on the nature of the roots. Let us take something more hands on: If
$$
x^2 + px + q = 0,
$$
what is $x$?

 

[Orodruin]

Sorry, I am not thinking straight. The expression you have is not the inner product but the speed at time t. How is the inner product of two vectors defined?

 

[throneoo]

Mod note: I have deleted the content of this post, as it was a complete solution.

 

[throneoo]

just realized i have made the mistake of not making B negative (since the velocity should be downwards)

so the expression should be t=(u/gsinA)

 

[Orodruin]

Let B be the angle of velocity in mid-air with respect to the horizon, such that the velocity is perpendicular to its original velocity.

A=pi/2-B

This expression is not correct as it gives A+B = pi/2 which is not equivalent to orthogonality. Also, note that forum rules forbids posting full solutions until the OP has solved the problem.

 

[throneoo]

doesn't that make the angle between the velocities orthogonal?

 

[throneoo]

vector approach:
u.v=0

(ucosA i + usinA j).(vcos(-B) i + vsin(-B) j )=0

(ucosA i + usinA j).(ucos(A) i + (usinA-gt) j )=0

 

[Orodruin]

No, in order to be orthogonal you would need A-B = pi/2. That B is negative is being taken care of automatically by the vector component.

 

[throneoo]

No, in order to be orthogonal you would need A-B = pi/2. That B is negative is being taken care of automatically by the vector component.

yup I realized that mistake

 

[Yashbhatt]

A)

I used this but got an expression I couldn't solve.

 

[Yashbhatt]

vector approach:
u.v=0

(ucosA i + usinA j).(vcos(-B) i + vsin(-B) j )=0

(ucosA i + usinA j).(ucos(A) i + (usinA-gt) j )=0

I used this, but could not solve the expression I got.

 

[Orodruin]

If you do the inner product equal to zero approach you should get a linear expression in t. It should be a simple matter of solving this.

Note that what you had earlier was not the inner product of the velocity at t and that at time 0, but the speed at time t. That will never be zero unless the horizontal component is.

 

[Yashbhatt]

(ucosA i + usinA j).(ucos(A) i + (usinA-gt) j )=0

I have to this, right?

 

[Orodruin]

Yes, so what do you get when you do that?

 

[throneoo]

(ucosA i + usinA j).(ucos(A) i + (usinA-gt) j )=0

I have to this, right?

 

[Yashbhatt]

I got u/g sin A. Is that correct?

 

[Orodruin]

I got u/g sin A. Is that correct?

Yes. The inner product becomes
$$
\vec v(0) \cdot \vec v(t) = u^2 - ugt\sin\theta = 0.
$$
Solving for $t$ gives the answer you just gave.

Note that there is no solution if $\theta=0$ and that the solution for $\theta<0$ is negative, meaning that this would have occurred earlier along the parabola and will not occur after the release.

Edit: Another special case is when $\sin\theta = 1$, i.e., the motion is purely vertical. Can you explain the result for this case?

 

[Yashbhatt]

I actually discussed this with my friend. That would give you u/g i.e. the highest point. At the highest point, the velocity will be a null vector. So, we can take it as perpendicular to the g vector.

Is that correct?

 

[Orodruin]

Yes, if the motion is purely vertical, then the velocity on top is zero and the time to get there is just the time u/g which is how long it takes to decelerate u to 0 with deceleration g. The inner product of the zero vector with anything is zero so it is orthogonal to everything.

 

[Yashbhatt]

Okay. Got it. Thanks for helping.

Can we find this for any arbitrary angle A?