As you know, the trajectory of a body kicked with velocity v at angle θ is
y = x tanθ - g/(2 (v cosθ)^2) x^2
For your ball, that is
y(x) = 1.11x - 9.81/(2(20*0.67)^2) x^2
= 1.11x - 0.0273x^2
So, now check to see whether y(33) >= 3.05
A place kicker must kick a football from a point 33.0 m (about 36 yd) from the goal. As a result of the kick, the ball must clear the crossbar, which is 3.05 m high. When kicked, the ball leaves the ground with a speed of 20.0 m/s at an angle of 48◦ to the horizontal.
To determine if the ball clears the crossbar, what is its height with respect to the crossbar when it reaches the plane of the crossbar? The acceleration of gravity is 9.81 m/s2 .
Answer in units of m
1 answer