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A projectile is fired from the surface of the earth with a velocity of 5ms-1 and angle \theta with the horizontal. Another projectile fired another planet with a velocity of 3 ms-1 at the same angle follows a trajectory which is identical with the trajectory of the projectile fired from the earth. The value of the acceleration due to gravity on the planet is (in ms-2) is: (given g = 9.8 ms-2)

  • Option 1)

    3.5

  • Option 2)

    5.9

  • Option 3)

    16.3

  • Option 4)

    110.8

 

Answers (2)

best_answer

As we learnt in

Horizontal Range -

Horizontal distance travelled by projectile from the point of projectile to the point on ground where its hits.

R=frac{u^{2}sin 2Theta }{g}

 

 

 

- wherein

Special case of horizontal range

For man horizontal range.

Theta = 45^{0}

R_{max}=frac{u^{2}sin 2 (45) }{g}=frac{u^{2} imes 1}{g}=frac{u^{2}}{g}

 

 

 

Range should be equal            

\frac{u_{e}^{2}\sin 2\theta}{2g_{e}}  =\frac{u_{p}^{2}\sin2\theta }{2g_{p}}                              (P \rightarrow plant,          e \rightarrow earth)         

\frac{5^{2}\times \sin 2\theta }{2\times 9.8} = \frac{3^{2}\times \sin 2\theta }{2g_{p}}

g_{p}=\frac{9.8\times 9}{25}=3.52\:\: m/s^{2}

Correct option is 1.


Option 1)

3.5

Correct

Option 2)

5.9

Incorrect

Option 3)

16.3

Incorrect

Option 4)

110.8

Incorrect

Posted by

prateek

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2

Posted by

Satyam Chaudhary

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