Question

Pls solve this question​

Answer 1

Answer:

the answer is square root of 18

Explanation;

For particle 1:

u=10m/s

g=-10m/$s^{2}$

to find the time in which the particle reaches the highest point at which v=0

v=u+at

0=10-10t

t=1

Highest point of particle 1 at t=1

s=10*1-1/2(10)1*1

s=10-5

s=5m

For particle 2:

ux=10cos(30)=5 $\sqrt{3}$

uy=10sin(30)=5

in time 1 second when the particle 1 is closest to particle 2 the x position of particle 2 will be

s=ut+1/2(a)$t^{2}$

s=5 $\sqrt{3}$+1/2(0)1*1 [a=0 since there is no  acceleration in x coordinate]

s=5 $\sqrt{3}$

in time 1 second when the particle 1 is closest to particle 2 the x position of particle 2 will be

s=ut+1/2(a)$t^{2}$

s=5-1/2(10)1*1

s=0

so when the particle 1 is closest to particle 2 the positions of particle 1=5 m above the ground particle 2=6$\sqrt{3}$ above the ground that implies particle 2 is 6

On solving we get

$\sqrt{(5\sqrt{3})^{2}+(6\sqrt(3)-5)^{2}}$

$\sqrt{15+18-60\sqrt{3}+25$

$\sqrt{58-60\sqrt{3}}$

The Answer seems to be an imaginary number and since length cannot be an imaginary number my conclusion is that your question is wrong