Question

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2 particles are projected vertically upwards from points (0,0) & (1,0) with uniform velocity 10 m/s & V m/s respectively, as shown in fig. It is found that they collide after time 't'. Find 'v' & 't'.

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Answer 1

To find:-

• Velocity and time

Solution:-

Let's assume that at time T they will collide.

Also when they will collide they will be at same height so we can find their height using end law of motion.

according to first ball the height at which they collide will be

$\large \mathtt{h = u_yt + \frac{1}{2}a_y {t}^{2} }$

$\large \mathtt{ u_y = 10 \sin \theta }$

$\large \mathtt{a_y = - 10m {s}^{ - 2} }$

putting the values in formula

$\mathtt{h = 10 \sin {30}^{o} t + \frac{1}{2}( - 10) {t}^{2} }$

Now height according to second particle

$\mathtt{ u_y = v\sin \theta }$

$\mathtt{a_y = - 10m {s}^{ - 2} }$

Putting values in formula

$\mathtt{h = v\sin {45}^{o} t + \frac{1}{2}( - 10) {t}^{2} }$

Now accoridng to both particles H will be same hence we can compare both equations.

$\large \mathtt{10 \sin {30}^{o} t + \frac{1}{2} ( - 10) {t}^{2} = v\sin {45}^{o} t + \frac{1}{2}( - 10) {t}^{2} }$

$\large \mathtt{10 \sin {30}^{o}t = v\sin {45}^{o} t }$

$\large \mathtt{10 \sin {30}^{o} = v\sin {45}^{o} }$

From here

$\large \mathtt{5 \sqrt{2} = v }$

Now the distance between the particles ground position is 1m hence the second particle will be traveled 1m less than the first particle when they will collide

Also there is no acceleration in horizontal component.

Hence we can write

$\mathtt{10 \ cos {30}^{o}t = x_{horizontal \: distance \: travelled \: by \: first \: ball} }$

$\mathtt{v \ cos {45}^{o} t= y_{horizontal \: distance \: travelled \: by\: second \: ball} }$

$\mathtt{also \: \: \: x = y + 1 }$

Hence

$\mathtt{10 \ cos {30}^{o} t= v \cos {45}^{o}t + 1 }$

putting value of V

$\large \mathtt{10 \ cos {30}^{o} t= 5 \sqrt{2} \cos {45}^{o}t + 1 }$

$\large \mathtt{10 \ cos {30}^{o} t - 5 \sqrt{2} \cos {45}^{o}t= 1 }$

$\large \mathtt{5 \sqrt{3} t - 5 t= 1 }$

$\large \mathtt{t(5 \sqrt{3} - 5)= 1 }$

$\large \mathtt{t= \frac{1}{(5 \sqrt{3} - 5)} }$

Answer

• $\large \mathtt{t= \frac{1}{(5 \sqrt{3} - 5)} }$

• $\large \mathtt{5 \sqrt{2} = v }$

Answer 2

Considering motion in horizontal direction,time taken to collide,

$t = 1 10√(3)2- v√(2)t = 2√(2)10√(6)-2v ...(1)In vertical direction, 10</p><p></p><p>$

$\sf{In vertical direction,}  \\ 10×\frac{1}{2} t− \frac{1}{2} gt = \sqrt[v]{12} \times$

$\sf - \frac{1}{2} gt {}^{2}$

$= \sqrt[v]{12}$

$= 5 \sf \sqrt []{2}$

$=2 \: \sqrt[]{2}$

_____________________

$5 \sqrt{6} - 10 \sqrt{6}$

$\sf{ =1.49 \: seconds \: is \: your \: answer}$

I hope this helps ☺️

if this is wrong I am sorry