A place kicker must kick the football from a point [06] m from the goal and clear a bar 3.00 m above the ground. The ball leaves the ground with a speed of 20.0 m/s at an angle of 53.0 degrees above the horizontal. (a) By how much (m) does the ball clear (positive value) or fall short (negative value) of the cross bar? (This is vertical distance above or below the cross bar.) (b) When it gets to the cross bar, what is the vertical component of the ball’s velocity (m/s)? (Is it rising or falling-pay attention to the sign?)

Explanation:

Horizontal component of initial velocity of throw = 20 cos 53 = 12 m /s

Vertical component = 20 sin 53 = 15.97 m /s

Distance to be travelled horizontally = 6 m .

time taken by ball to travel this distance = 6 / 12 = 0.5 s

vertical displacement during this period can be calculated as follows .

Initial vertical velocity = 15.97 m /s

time of travel = .5 s

acceleration = - 9.8 m /s²

s = ut - 1/2 g t²

= 15.97 x .5 - .5 x 9.8 x 0.5²

= 7.985 - 1.225

= 6.76 m

Goal post is 6 m high , so ball will cross the goal post .76 m or 76 cm above cross bar  .

b ) vertical component of ball when it crosses the goal post . Let it be v .

v = u - gt

Applying this formula for vertical movement ,

v = 15.97 - 9.8 x .5

= 15.97 - 4.9

= 11.07 m /s .

i guess it’s a and b

frequency or wavelenth of em radiation in a vacuum