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Derive the Formula of Projectile Motion!

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Answer 1

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Answer 2

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Let us consider a ball projected at an angle θ with respect to horizontal x-axis with the initial velocity u as shown below:

Total Time of Flight: Resultant displacement (s) = 0 in Vertical direction. Therefore, by using the Equation of motion:

gt2 = 2(uyt – sy) [Here, uy = u sin θ and sy = 0]

i.e. gt2 = 2t × u sin θ

Therefore, the total time of flight (t):

Horizontal Range: Horizontal Range (OA) = Horizontal component of velocity (ux) × Total Flight Time (t)

R = u cos θ × 2u×sinθg

Therefore in a projectile motion the Horizontal Range is given by (R):

Maximum Height: It is the highest point of the trajectory (point A). When the ball is at point A, the vertical component of the velocity will be zero. i.e. 0 = (u sin θ)2 – 2g Hmax [s = Hmax , v = 0 and u = u sin θ]

Therefore in a projectile motion the Maximum Height is given by (Hmax):

The equation of Trajectory: Let, the position of the ball at any instant (t) be M (x, y). Now, from Equations of Motion:

x = t × u cos θ . . . . . . (1)

y = u sin θ × t – 12×t2g. . . . . . (2)

On substituting Equation (1) in Equation (2):

This is the Equation of Trajectory in a projectile motion, and it proves that the projectile motion is always parabolic in nature.