What is the angle of projection to have a maximum range in kitti pull?if one strikes kitti pull with of 98ms_1 what is the maximum range achieved
Answers
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.
Hope it helps.
Answer:
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