What is the angle of projection to have a maximum range in kitti pull?if one strike kitti pull with the speed of 98ms–1 what is the maximum range achieved?
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The maximum angle of the Projection for having the maximum range is 45°.
Reason ⇒ As we all know that the maximum m value of the Sine is 1 which is on 90°.
∴ From the Formula of Range, R = u²Sin2θ/g,
we must need to multiply 2 by 45° to get 90°.
This can give the maximum angle for maximum range. Initial velocity can also increase to give maximum range.
Now, For the Value of the Maximum Range,
R = 98² × 1/9.8
= 98 × 98/9.8
= 980 m.
∴ Maximum Range of the Kitti Pull is 980 m.
Hope it helps.
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Heyy mate! View the picture which i had posted
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