$H=(u^2 sin^2\theta )/(2g)$

$h-H =(1/2)gt^2 = (u^2sin^2 \theta)/(2g)$

From this I got h=2H

But answer is 4H

1 vote

Best answer

**Revised Answer**

If the particles are launched at the same time, they must have the same *horizontal speed* in order to collide. (However, the problem seems to state that the particles have the same speed, ie same magnitude of velocity, and it is not clear that they are launched at the same time.)

Because they have the same horizontal speed, they remain aligned vertically. We can ignore horizontal motion and treat this as motion in 1D vertically.

The particles must collide before they reach the ground. The maximum $h$ occurs when they collide at the ground. ie They reach the ground at the same time, and have the same time of flight $t$.

When the upper particle reaches the ground $h=\frac12gt^2$. When the lower particle falls from its maximum height it takes time $\frac12 t$ so

$H=\frac12g(\frac12 t)^2=\frac14(\frac12gt^2)=\frac14h$

$h=4H$.

+1 Actually your both answers are correct.

2 votes

Speed of paricle 1 be u at an angle $\theta$ from ground .Then the speed of particle 2 is $u\cos \theta$ in horizontal direction.

let at any h' from ground both partucles collide .

For particle 1 $$h'= u\sin\theta t -(1/2)gt^2$$

For particle 2 $$h-h '= (1/2) gt^2$$

Hence $$t=\sqrt {\frac{2(h-h')}{g}}$$

Also we know $H=u^2 sin\theta ^2 /(2g)$

Substituting these values in equation of particle 1

We geta quadratic equation $$h^2 -4Hh+4Hh'=0$$

solving this $$h=\frac{4H(+/-)\sqrt{16H^2 -16Hh'}}{2}$$

Now h will be maximum when h'=0 that is they collide when both of them just going to strike the ground .

$$h=\frac{4H+4H}{2}$$

$$h=4H$$

1 vote

Your Equation considers the particles to collide only at the point where first particle attains its maximum height. But the question considers the general case.

Since they are thrown from the same vertical plane with the same speed, they can collide only if the horizontal component of their velocities are equal. Here the second particle would have covered a large distance when they are at the same height.

If the first particle is necessarily thrown obliquely then the particles cannot collide I assume.

1 vote

**Discussion**

DoubtExpert correctly observes that the particles do not necessarily collide when the particle thrown obliquely reaches its maximum height. However, he *assumes* that the particles are launched simultaneously. I think this is not required by the wording of the question.

The key phrase is *thrown to strike at same time*. This is ambiguous. If this means (as I think it does) that the particles collide at the same time then it is superfluous, because they must be in the same place at the same time in order to collide. I do not see how it can be interpreted as *launched at the same time.* If it does mean this then DoubtExpert is correct : there can be no collision.

**Solution**

We must assume that the particles can be launched independently, at any time. The only requirement is that the trajectories overlap - or at least touch tangentially - at some point in space. If this happens then we can always adjust the launch times to ensure that the particles collide at this point.

What we need to do is write equations for the two trajectories then find a condition for them to touch. The trajectories will touch if (i) they intersect at some point, and (ii) their tangents are equal at this point.

Suppose the speed of each particle is $u$ and the launch angle of the lower particle is $\theta$. Writing $T=\tan\theta$ the trajectories of the upper and lower particles are respectively

$y=h-\frac{g}{2u^2}x^2$

$y=xT-\frac{g(1+T^2)}{2u^2}x^2$.

The tangents have slope

$\frac{dy}{dx}=-\frac{g}{u^2}x$

$\frac{dy}{dx}=T-\frac{g(1+T^2)}{u^2}x$.

The 1st two equations are equal when

$h=xT-\frac{gT^2}{2u^2}x^2$.

The 2nd two equations are equal when

$T=\frac{gT^2}{u^2}x$

$xT=\frac{u^2}{g}$.

Combine these two results :

$h=\frac{u^2}{2g}$.

The height of the collision point is

$y=h-\frac{g}{2u^2}x^2=\frac{u^2}{2g}-\frac{u^2}{2gT^2}=h(1-\frac{1}{T^2})$.

This must be above ground $(y \ge 0)$ so

$T^2 \ge 1$

$\theta \ge 45^{\circ}$.

The maximum height reached by the lower particle is $H=\frac{u^2\sin^2\theta}{2g}$. So we have $\frac{H}{h}=\sin^2\theta$. The value of $h$ calculated here is the maximum possible - any larger and the trajectories would not intersect. So the criterion is

$h \ge \frac{H}{\sin^2\theta}$.

I do not think we can make any further progress without knowing the launch angle $\theta$. In order to get $h = 4H$ we require $\theta=30^{\circ}$. However, this is not $\ge 45^{\circ}$ : the collision in this case takes place below ground. So a collision is not possible when $h =4H$.

If the particles are launched with the same speed at the same time, then I agree with DoubtExpert : the particles cannot collide.

Is there a worked solution? Or a diagram?

Is there a worked solution? Or a diagram?

Hmm, true because horizontal distance being same $ut = u cos \theta t $

and $\theta = 0 $ so this may not cause collision or because even the speed given are same.

and $\theta = 0 $ so this may not cause collision or because even the speed given are same.

The solution has bad explanation .

http://pasteboard.co/MmmDW5wny.jpg

http://pasteboard.co/MmmDW5wny.jpg

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