Aball is thrown with a velocity of 20 m/s at an angle of 60 degrees above the horizontal. what is the horizontal component of its instantaneous velocity at the the exact top of its trajectory?

10 m/s.

Explanation:

Assuming no air resistance, horizontal component of its velocity will be constant throughout the motion of 10 m/s.

Formula is,

Let horizontal component of velocity be hv.

Now,

cos( 60 ) = hv / 20.

hv = 20 * cos ( 20 )

hv = 10 m/s.

r = ρ x length / area  

ρ = r x area / length  

ρ is in ohm-meters  

in basic units, it is m³kg/s³a²  

i think this is it.