Final answer:
The ball was thrown at an angle of approximately 30.0° using the given values for the range and initial speed. The other angle that gives the same range is 60 degrees, but it is not used because the problem specifies using the smaller angle. The pass took approximately 0.808 seconds to complete.
Step-by-step explanation:
(a) To determine the angle at which the ball was thrown, we can use the equation for the horizontal range of a projectile. The range is given by the equation:
R = (v0)2sin(2θ) / g
where R is the range, v0 is the initial speed of the ball, θ is the angle of projection, and g is the acceleration due to gravity. We are given the range (R = 7.00 m) and the initial speed (v0 = 12.0 m/s). Rearranging the equation to solve for θ, we have:
θ = sin-1((Rg) / (v0)2) / 2
Substituting the given values into this equation:
θ = sin-1((7.00 m)(9.8 m/s2)) / (12.0 m/s)2) / 2
θ ≈ 30.0°
(b) To find the other angle that gives the same range, we can use the fact that the range is symmetrical for two angles that add up to 90 degrees. In this case, the other angle would be 90 - 30 = 60 degrees. However, this angle would not be used because the problem states that the smaller of the two possible angles was used.
(c) To calculate the time it took for the pass, we can use the equation:
t = R / (v0cos(θ))
where t is the time, R is the range, v0 is the initial speed, and θ is the angle of projection. Substituting the given values:
t = (7.00 m) / ((12.0 m/s)cos(30.0°))
t ≈ 0.808 s