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When a stone is projected, its horizontal range is 24m and greatest height 6m. Find its velocity of projection.​

User Nanhydrin
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1 Answer

1 vote

Answer:

15.33 m/s

Step-by-step explanation:

We are here given that ,

  • Range = 24m (R)
  • Maximum height= 6m (h)
  • velocity of projection= ? (u)

As we know that,


\longrightarrow R = (u^2\sin2\theta)/(g) \dots (1)\\

And ,


\longrightarrow h =(u^2\sin^2\theta)/(2g) \dots (2)\\

Divide equation 1 and 2 ,


\longrightarrow (R)/(h)=(4 \cos\theta)/(\sin\theta) \\


\longrightarrow (24m)/(6m)=4 \cot\theta \\


\longrightarrow \cot\theta = 1 \\


\longrightarrow \theta = \cot^(-1)(1) \\


\longrightarrow \underline{\underline{\theta = 45^(\circ)}}\\

Now we may substitute this value in equation 1 as ,


\longrightarrow 24 =( u^2\sin(2* 45^(\circ) ))/(g) \\


\longrightarrow 24g = u^2\sin90^\circ \\


\longrightarrow u^2 = 24 * 9.8 \\


\longrightarrow u =√(235.2) m/s \\


\longrightarrow \underline{\underline{ u \approx 15.33 \ m/s }}\\

and we are done!

User Tomaso Albinoni
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