A place kicker must kick a football from a point 35.0 m (= 38.3 yd) from the goa

Question

A place kicker must kick a football from a point 35.0 m (= 38.3 yd) from the goal, and the ball must clear the crossbar, which is 3.05 m high. When kicked, the ball leaves the ground with a speed of 21.3 m/s at an angle of 54.0o to the horizontal. By how much does the ball clear the crossbar (if in fact it does)? Enter positive values if the ball clears and negative values if it falls below the crossbar. What is the vertical velocity of the ball at the time it reaches the crossbar? Enter positive values if it is still rising and negative values if it is falling. A place kicker must kick a football from a point 35.0 m (= 38.3 yd) from the goal, and the ball must clear the crossbar, which is 3.05 m high. When kicked, the ball leaves the ground with a speed of 21.3 m/s at an angle of 54.0o to the horizontal. By how much does the ball clear the crossbar (if in fact it does)? Enter positive values if the ball clears and negative values if it falls below the crossbar. What is the vertical velocity of the ball at the time it reaches the crossbar? Enter positive values if it is still rising and negative values if it is falling. A place kicker must kick a football from a point 35.0 m (= 38.3 yd) from the goal, and the ball must clear the crossbar, which is 3.05 m high. When kicked, the ball leaves the ground with a speed of 21.3 m/s at an angle of 54.0o to the horizontal. By how much does the ball clear the crossbar (if in fact it does)? Enter positive values if the ball clears and negative values if it falls below the crossbar. What is the vertical velocity of the ball at the time it reaches the crossbar? Enter positive values if it is still rising and negative values if it is falling.

Explanation / Answer

we wlil use projectile motion equations,

Ux= horizontal component of velocity = 21.3 cos 54 = 12.52 m/s apprx

time to reach the cross bar= 35 m / 12.52= 2.795 seconds apprx

h= ( vertical distance travlled during this time) = 21.3 sin 54 ( 2.795) - 1/2 (9.8) ( 2.795)^2=9.884 m apprx

ball will clear the cross bar = ( 9.884 - 3.05) = 6.834 m apprx

vertical component of velocity on reaching the cross bar = 21.3 sin 54 - 9.8 (2.795 ) =17.232 - 27.391= -10.159 ( it's going down)

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