Question

Derive the relation between maximum range and maximum height

Answer 1

Let a particle is projected with speed u m/s at an oblique angle Θ with respect to horizontal direction. As we know, in projectile motion , acceleration on horizontal direction is zero and acceleration on vertical direction is -g .

Ux = ucosΘ

Uy = usinΘ

Let after T time , particle complete the motion.

so, net displacement in vertical direction after T time = 0

Now, use formula , S = ut + 1/2at² for Y - axis .

Here, S = 0, u = usinΘT - 1/2gT²

T = 2usinΘ/g

Hence, time of flight , T = 2sinΘ/g

Now, Range , R = UxT = ucosΘ.2usinΘ/g

Hence, Range (R) = u²sin2Θ/g

at maximum height , only horizontal component of velocity is appear.and time taken to reach equals half of time of flight.

V² = U² + 2aS

Vy² = Uy² + 2ayH

0 = u²sin²Θ - 2gH

H = u²sin²Θ/2g

Hence, maximum height , H = u²sin²Θ/2g

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Answer 2

heya refee to the attachemt please