Question

Derive the equation of motion, range and maximum height reached by the particle thrown at an oblique angle Θ with respect to the horizontal direction.

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

Answer 2

First, there are some essential equation regarding one-dimension motion with constant acceleration:

v=v0+atv=v0+at (1)

x=x0+v0t+12at2x=x0+v0t+12at2 (2)

v2−v20=2asv2−v02=2as (3)

(where v is the final velocity, v0v0 is the initial velocity, x is the displacement, s is the distance the object travels along the direction of acceleration)

(All this set of equations can be proved with simple calculus)

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