What If? If the teacher always launches the ball with the speed found in part (a), what is the minimum angle (in degrees above the horizontal) at which he can launch the ball and still clear the playground railing? (Hint: You may need to use the trigonometric identity sec2(θ) = 1 + tan2(θ).)

9514 1404 393

  43.303°

Step-by-step explanation:

Problem geometry varies. For this question, we will assume ...

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.

hi there!

step-by-step explanation:

a roof pitch is basically for every inch rise there is 12 inches run.

the height of the roof is also known as the rafter.

formula: rafter^2 = rise^2 + run^2

rafter^2 = 5^2 + 12^2

rafter^2 = 25 + 144

rafter ^2 = 169"^2

the height of the roof from the lowest point to the highest point is 169"^2.

hope this : )