9514 1404 393
Answer:
43.303°
Step-by-step explanation:
Problem geometry varies. For this question, we will assume ...
- The ball is launched at a 53° angle, 24 m from a building
- The ball takes 2.20 seconds to reach the building
- The railing is 6.4 m high
The equation of motion in this case is ...
h(x, α) = x(tan(α) -4.9x/(v·cos(α))²)
where h(x) = height at distance x, α is the launch angle, vₓ is the horizontal speed of the ball.
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The ball was launched at an angle of 53° in part (a) at a speed that gives it a horizontal speed of (24 m)/(2.2 s). That is, the speed in the direction of launch is ...
(24 m)/(2.2 s)/cos(53°) ≈ 18.126983 m/s . . . . speed in part (a)
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For the question at hand, we need to find α such that ...
h(24) = 6.4
Substituting into the above equation gives ...
6.4 = 24(tan(α) -24(4.9/18.126983^2·sec(α)^2)
6.4 = 24(tan(α) -0.35789551(tan(α)^2 +1)) . . . . replacing sec(α) using hint
If we let z represent tan(α), this can be written in standard form as ...
8.5894922·z^2 -24z +14.989492 = 0
Then the minimum solution (by the quadratic formula) is ...
z = (24 -√(24^2 -4(8.5894922)(14.989492)))/(2·8.5894922)
z = 0.94244774
And the angle is ...
α = arctan(z) = arctan(0.94244774) ≈ 43.303°
The teacher must launch the ball at an angle of at least 43.3° to clear the railing.
