Question

A body of mass m is thrown with velocity v at angle of 30 degrees to th horizontal and another body b of same mass is thrown with velocity v at an angle of 60 degrees to the horizontal. find the ratio of the horizontal range and maximum height of a and b?

Answer 1

See the attachment for your answer...

Hope it helps..... :-)

Answer 2

Answer: Hᵃ: Hᵇ=$1:\sqrt{3}$ and   Rᵃ: Rᵇ=1:1.

Step-by-step explanation:

• To find the maximum height (H)=$\frac{v^{2}sin^{2}\alpha }{2g}$
• To find the range (R)=$\frac{v^{2}sin{2}\alpha }{g}$
• Case-1 [$\alpha =30$]

       Hᵃ=$\frac{v^{2}sin^{2} 30 }{2g}$

           =$\frac{v^{2}(\frac{1}{2} )^{2} }{2g}$

           =$\frac{v^{2} }{8g}$

     Rᵃ=$\frac{v^{2}sin{60} }{g}$

         =$\frac{\sqrt{3} v^{2} }{2g}$

• Case-2 [ $\alpha =60$]

       Hᵇ=$\frac{v^{2}sin^{2} 60 }{2g}$

           =$\frac{v^{2}(\frac{\sqrt{3} }{2} )^{2} }{2g}$

           =$\frac{\sqrt{3} v^{2} }{8g}$

       Rᵇ=$\frac{v^{2}sin{120} }{g}$

          =$\frac{\sqrt{3} v^{2} }{2g}$

• Now clearly, Hᵃ: Hᵇ=$1:\sqrt{3}$ and   Rᵃ: Rᵇ=1:1.
• Hence,the required ratio of maximum height and horizontal range is $1:\sqrt{3}$ and 1:1.

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