Arugby player passes the ball 6.78 m across the field, where it is caught at the same height as it left his hand. (assume the player is facing the +x direction.) (a) at what angle (in degrees counterclockwise from the +x-axis) was the ball thrown if its initial speed was 13.1 m/s, assuming that the smaller of the two possible angles was used? ° counterclockwise from the +x-axis (b) what other angle (in degrees counterclockwise from the +x-axis) gives the same range? ° counterclockwise from the +x-axis why would it not be used? the larger angle would provide a smaller maximum height, reducing the chances that the receiver would see the ball mid-flight. the larger angle would provide a smaller maximum height, increasing the chances that an opponent would intercept the ball mid-flight. the larger angle would give the receiver less time to achieve an optimal offensive position. the larger angle would give an opponent more time to achieve an optimal defensive position.

a) The ball was thrown at an angle of 11.4°

b) The other larger angle that gives the same range is 348.6°. This angle will not be used because the ball would need to have an acceleration that is 9.8 m/s² upwards. But supposing it does, then the maximum height will be smaller increasing the chances of the opponents to catch the ball.

Explanation:

Please, see figure 1 for a description of the problem

The position of the ball is given by the vector position "r":

r = (x0 + v0 * t * cos θ, y0 + v0 * t * sin θ + 1/2 g * t²)

where

r = vector position

x0 = initial horizontal position

t = time

θ = throwing angle

y0 = initial vertical position

v0 = initial speed

g = acceleration due to gravity (-9.8 m/s²)

Since the origin of the reference system is the point at which the ball is thrown, x0 and y0 = 0.

a) From the figure, we know that when the position vector is "r final", the x-component of the vector is the horizontal distance traveled by the ball and the y-component is 0.

Then:

v0 * t * cos θ = 6.78 m

v0 * t * sin θ + 1/2 g * t² = 0

We have 2 equations with 2 variables (θ and t), then, we can solve the system.

Solving for t in the second equation:

t (v0 * sin θ + 1/2 g * t) = 0

t = 0

and

v0 * sin θ + 1/2 g * t = 0

t = -2 * v0 sin θ / g = -2 * 13.1 m/s * sin θ /(-9.8 m/s ²)

t = 2.67 s * sin θ

Replacing "t" in the first equation:

13.1 m/s * (2.67 s * sin θ) * cos θ = 6.78 m

35.0 m * sin θ * cos θ = 6.78 m

sin θ * cos θ = 0.194

sin (2θ) / 2 = 0.194

sin 2θ = 0.388

2θ =22.8

θ = 11.4°

b) The other angle that gives the same range is 360°- θ = 348.6°

Because cos 11.4° = cos 348.6°

See the figure 2 for the trajectory the ball would have with this angle. This trajectory is impossible because the ball would have to be accelerated 9.8 m/s² upwards. But let´s assume that by some magic effect, the ball has that trajectory. Then, the maximum height will be smaller, increasing the chances of the opponents to intercept it. Although they would not expect that!

here we know that ball will drop down by total distance 3.2 m

so here we know that

now using kinematics

now in the same time it will cover horizontal distance of 0.75 m

so speed of the ball will be

electrical power