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A rock is thrown upward from level ground in such a way that the maximum height of its flight is equal to its horizontal range R.

AAt what angle theta is the rock thrown?

BIn terms of its original range R, what is the range R_max the rock can attain if it is launched at the same speed but at the optimal angle for maximum range?

CWould your answer to part (a) be different?

Verified Answer :

the range equation is R=v0^2 sin(2t)/g where I use t for theta 

so, take the ratio of Rmax to R: 

Rmax/R = v0^2 sin(2*45)/g / v0^2sin(2t)/g 

the v0 and g terms cancel, sin 90=1, leaving 

Rmax/R = 1/sin(2t) or Rmax = R/sin(2t)

that would help

olayemiolakunle65

Answer:

(a) Θ = \(75.96^{0}\)

(b) \(R_{max}\) is achieved when T = \(45^{0}\), thus \(R_{max}\) = R ÷ Sin2T

(c) Yes

Explanation:

(a)  Max. height (\(H_{max}\)) = \(\frac{U^{2}Sin^{2}T}{2g}\)

     Range (R) = \(\frac{U^{2}Sin 2T }{g}\)

where T represents theta

So for, R = \(H_{max}\)

⇒  \(\frac{U^{2}Sin 2T }{g}\) = \(\frac{U^{2}Sin^{2}T}{2g}\)

Using trigonometric functions,

        \(Sin^{2}T\) = 2SinTCosT = \(\frac{Sin^{2}T }{2}\)

  ⇒ 4 = Tan T

        T = \(Tan^{-1} 4\)

           = \(75.96^{0}\)

∴ Theta = \(75.96^{0}\)

(b) Since,  R = \(\frac{U^{2}Sin 2T }{g}\)

\(R_{max}\) is achieved when T = \(45^{0}\)

\(R_{max}\) = \(\frac{U^{2}Sin 2T }{g}\)

∴   \(R_{max}\)   = R ÷ Sin2T

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