Question

From the same point, two balls A and B are thrown simultaneously. A is thrown vertically up with a velocity of 20 m/s. B is thrown with a velocity of 20 m/s at an angle of 60° with the vertical. Determine the separation between the two balls at t=1s

Answer 1

Answer:

Watch the explanation....

Explanation:

At first... the position of A is vertically up.....

$h = 20(1) - 4.9( {1}^{2} )$

so, h= 15.1m

Now for the B, we get position in x component at time to 1 .....

$x = 20 \cos(30) (1)$

here the B is angled 60° with y component and 30° with x component.

so, x = 10root3 m.

Now, for y component we get.....

$y = 20( \sin(30) ) - 4.9( {1}^{2})$

and, y = 5.1 m

Now the horizontal distance between A and B is 10root3 or 17.3m

The vertical distance between A and B is 10 m

And the all round displacement is.....

$\sqrt{( {17.3)}^{2} + ( {10)}^{2} }$

which is about (correctly to say..) 20 m.

I am done.