Question

Derive an expression for (1) equation of trajectory (2) Time of flight (3) Maximum height attained (4) Horizontal range for the projectile fired at an angle with horizontal

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Answer 1

Maximum height (max): The maximum vertical distance travelled by the projectile during the journey is called maximum height. This is determined as follows: For the vertical part of the motion u 2 y uy2 = u 2 y uy2 + 2ays Here uy = usin0, a = -g, s = hmax, and at the maximum height v = 0 Time of flight (Tf): The total time taken by the projectile from the point of projection till it hits the horizontal plane is called time of flight. This time of flight is the time taken by the projectile to go from point O to B via point A as shown in figure. We know that Sy = uy + 1 2 12ayt2 Here, sy = y = 0 (net displacement in y-direction is zero), uy = u sin 0, a = -g, t = Tf , Then 0 = usin 0 Tf - 1 2 12g T 2 f Tf2 Tf = 2u s i n 0 g sin0g.

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Answer 2

Answer:

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Explanation:

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