Question

1) a football team plans for a field goal from the 45m line. crossbar is 3m above and it must clear it to be counted. he strikes it off the gorund at a velocity of 25m/s (30*)
A) will it clear the crossbar, explain?
B) what was the velocity when it pasts the post?

Answer 1

Answer on Question #64821, Physics / Other

1) a football team plans for a field goal from the 45m line. crossbar is 3m above and it must clear it to be counted. he strikes it off the ground at a velocity of $25\mathrm{m / s}$ ( $30{}^{\circ}$ )

A) will it clear the crossbar, explain?

B) what was the velocity when it pasts the post?

Solution:

Neglecting air resistance, the projectile is subject to a constant acceleration $g = 9.81 \, \text{m/s}^2$ , due to gravity, which is directed vertically downwards.

We use x-y coordinates with origin at the release point.

A)

For $x$ coordinate:

$$[ x = x _ {0} + (v _ {x}) _ {0} t ]$$ $$4 5 = 0 + (2 5 \cos 3 0 {}^ {\circ}) t$$

We now find the flight time

$$t = \frac {4 5}{2 5 \cos 3 0 {}^ {\circ}} = 2. 0 7 8 \mathrm {s}$$

For y coordinate:

$$\left[ y = y _ {0} + (v _ {y}) _ {0} t - \frac {1}{2} g t ^ {2} \right]$$ $$y = 0 + (2 5 \sin 3 0 {}^ {\circ}) \times 2. 0 7 8 - \frac {9 . 8 1}{2} (2. 0 7 8) ^ {2} = 4. 8 \mathrm {m}$$

The ball will clear the crossbar, because $y > h$ .

B)

The kinematic equation that describes an object's motion in vertical direction is:

$$v _ {y} = v _ {o y} - g t = (2 5 \sin 3 0 {}^ {\circ}) - 9. 8 1 \cdot 2. 0 7 8 = - 7. 9 \mathrm {m / s}$$

The horizontal component of velocity is $25\cos 30{}^{\circ} = 21.65\mathrm{m / s}$ m/s and the vertical component of velocity is $-7.9\mathrm{m / s}$ .

The final speed is

$$v _ {f} = \sqrt {v _ {x} ^ {2} + v _ {y} ^ {2}} = \sqrt {21.65^{2} + 7.9^{2}} \approx 23.0 \mathrm{m/s}$$

Answer: A) The ball will clear the crossbar, because $4.8\mathrm{m} > 3\mathrm{m}$; B) $23.0\mathrm{m/s}$.

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