Question

a football goal crossbar is 10 feet high.suppose a football placed on the field is 30 yards horizontally from a goal crossbar. a football player wants to kick the ball to score .discuss what initial velocity magnitude in si unit and initial velocity degree angle to the horizontal you would use so that the ball just clears the crossbar .show your work ( note 3.28 feet=1 meter.3feet=1 yard.) use si unit only please.please explain why each velocity and angle is chosen.​

Answer 1

Given:

Height of the goal crossbar $=10ft$

Distance of the football from the goal bar $=30yards$

$1m=3.28ft$

$1yard=3ft$

To find:

1. Initial velocity of projection.
2. Angle of projection.

Solution:

Step 1

We have been given that the football player is standing at a

distance of $30$ yards = $27.4$ meter from the goal crossbar, which is at a

height of $10$ feet = $3.048$ meter above the ground.

Let O be the point from which the football is kicked and the football follows a projectile motion till the goal is reached.

For getting a proper goal, the ball needs to be a bit lower than the height of the crossbar lets assume it to be at 3 meters.

Now,

The range of a projectile is given by $R=\frac{u^{2}sin2\alpha }{g}$

The height of a projectile is given by $H=\frac{u^{2} .sin^{2}\alpha }{2g}$

Where, $\alpha$ is the angle made by the projectile with the horizontal.

Step 2

Now, if we take the ratio of range and height, we get

$\frac{R}{H} =(\frac{u^{2}sin2\alpha }{g})(\frac{2g}{u^{2} sin^{2}\alpha } )$  

$\frac{R}{H} =\frac{2.sin2\alpha }{sin^{2}\alpha }$

$\frac{R}{H}=\frac{2(2sin\alpha .cos\alpha )}{sin^{2} \alpha }$

$\frac{R}{H} =4cot\alpha$

$4cot\alpha =\frac{27.4}{3}$

$cot\alpha =2.29$

$\alpha =23.5$°  $(approx)$

Hence, the angle with which the football needs to be kicked to get a goal with respect to the horizontal is 23.5°.

Step 3

Now,

We know, the height required for getting a goal is 3 m. Hence, we get

$\frac{u^{2} .sin^{2}\alpha }{2g}=3$

Substituting the values of $\alpha =23.5^{o}$ we get, we get $sin(23.5)=0.4$,

$\frac{u^{2} .(0.4)^{2} }{2(10)}=3$

$u^{2} =5\sqrt{15}$

$u=5$ × $3.8$

$u=19m/s$

Hence, the velocity with the ball needs to be kicked such that it gets into the goal is 19 m/s.

Final answer:

Hence,

1. The football needs to kicked with a speed of 19 m/s.
2. The football needs to be kicked at an angle of 23.5° with respect to the horizontal.