Answer:
a) The ball clears the crossbar by 10.6 m
b) The ball approaches the crossbar while falling
Step-by-step explanation:
The position of the ball is described by the vector position r (see attached figure):
r = (x0 + v0 t cos α ; y0 + v0 t sin α + 1/2 g t²)
where:
x0 = initial horizontal position
v0 = magnitude of the initial velocity
t = time
α = launching angle
y0 = initial vertical position
g = acceleration due to gravity (-9.8 m/s²)
The vector r is composed by rx and ry (see figure):
r = (rx ; ry)
a) Let´s find the time at which the ball flies a distance of 36.0 m. If at that time the vertical component of the vector r, ry, is equal or greater than 3.05 m, then the ball will clear the crossbar.
rx = x0 + v0 t cos α = 36.0 m
Since the origin of the system of reference is located where the kicker is, x0 = 0.
36.0 m = v0 t cos α
36.0 m /(v0 cos α) = t
36.0 m / (22.8 m/s * cos 51.0°) = t
t = 2.51 s
Now let´s calculate the height of the ball at that time:
ry = y0 + v0 t sin α + 1/2 g t²
Since the kicker is on the ground, y0 = 0
ry = 22.8 m/s * 2.51 s * sin 51.0° - 1/2 * 9.8 m/s² * (2.51 s)² = 13.6 m
Since the crossbar is 3.05 m high, the ball clears it by (13.6 m - 3.05 m) 10.6 m
b) Please see the figure to figure this out ;)
If the ball approaches the crossbar while still rising, the vertical component (vy) of the velocity vector will be positive. In change, if the ball approaches the crossbar while falling the vertical component of the velocity will be negative. See the figure.
The velocity vector is given by this equation:
v = (vx ; vy)
v = ( v0 cos α ; v0 sin α + g t)
Let´s see the vertical component at time t = 2.51
vy = v0 sin α + g t
vy = 22.8 m/s * sin 51.0° - 9.8 m/s² * 2.51 s
vy = -6.88 m/s
Then, the ball approaches the crossbar while falling.