A rugby 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 12.1 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)
(c) How long (in s) did this pass take?
To solve this problem, we can use the equations of projectile motion. Here's how we can find the answers to each part of the question:
(a) To find the angle at which the ball was thrown, we can use the range formula for projectile motion. The formula for the range (R) is given by:
R = (v^2 * sin(2θ)) / g
Where v is the initial speed of the ball, θ is the angle at which it was thrown, and g is the acceleration due to gravity (approximately 9.8 m/s^2).
Rearranging the formula to solve for θ, we get:
θ = 0.5 * arcsin((R * g) / v^2)
Substituting the given values into the formula, we have:
θ = 0.5 * arcsin((6.78 * 9.8) / (12.1^2))
Calculating this expression gives us the smaller angle at which the ball was thrown.
(b) To find the other angle that gives the same range, we can use the symmetry of projectile motion. Since the range is the same for two angles (θ and 90° - θ), we can subtract the angle calculated in part (a) from 90° to get the other angle.
For example, if the angle calculated in part (a) is 30°, then the other angle would be 90° - 30° = 60°.
(c) The time taken for the pass can be found using the equation:
R = v * cos(θ) * t
Where t is the time taken for the pass. Rearranging the equation, we have:
t = R / (v * cos(θ))
Substituting the given values, we get:
t = 6.78 / (12.1 * cos(θ))
Calculating this expression gives us the time taken for the pass.