Question

A rugby player passes the ball 7.00 m across the field, where it is caught at the same height as it left his hand.

(a) At what angle was the ball thrown if its initial speed was 12.0 m/ s, assuming that the smaller of the two possible angles was used?
(b) What other angle gives the same range, and why would it not be used?
(c) How long did this pass take?

Answer 1

a) θ = 14.23º, b) θ₂ = 75.77, c) t = 0.6019 s

Step-by-step explanation:

This is a missile throwing exercise.

a) the reach of the ball is the distance traveled for the same departure height

R =
`(v_o^2 \ sin 2 \theta )/(g)`

sin 2θ =
`(Rg)/(v_o^2)`

sin 2θ = 7.00 9.8 / 12.0²

2θ = sin⁻¹ (0.476389) = 28.45º

θ = 14.23º

the complementary angle that gives the same range is the angle after 45 that the same value is missing to reach 90º

θ ’= 90 -14.23

θ’= 75.77º

b) the two angles that give the same range are

θ₁ = 14.23

θ₂ = 75.77

the greater angle has a much greater height so the time of the movement is greater and has a greater chance of being intercepted by the other team.

C) the time of the pass can be calculated with the expression

x = v₀ₓ t

t = x / v₀ₓ

t = 7 / 11.63

t = 0.6019 s