show the path of projectile is para bolic
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Let a body is projected with speed u m/s inclined θ with horizontal line .
Then, vertical component of u, = ucosθ
Horizontal component of u , = usinθ
acceleration on horizontal, ax = 0
acceleration on vertical, ay = -g
Now, use formula ,
x =
x = ucosθ.t
t = x/ucosθ------(1)
Again, y =
y = usinθt - 1/2gt²
Put equation (1) here,
y = usinθ × x/ucosθ - 1/2g × x²/u²cos²θ
= tanθx - 1/2gx²/u²cos²θ
This equation is similar to Standard equation of parabola y = ax² + bx + c her, a, b and c are constant
So, A projectile motion is parabolic motion
Then, vertical component of u, = ucosθ
Horizontal component of u , = usinθ
acceleration on horizontal, ax = 0
acceleration on vertical, ay = -g
Now, use formula ,
x =
x = ucosθ.t
t = x/ucosθ------(1)
Again, y =
y = usinθt - 1/2gt²
Put equation (1) here,
y = usinθ × x/ucosθ - 1/2g × x²/u²cos²θ
= tanθx - 1/2gx²/u²cos²θ
This equation is similar to Standard equation of parabola y = ax² + bx + c her, a, b and c are constant
So, A projectile motion is parabolic motion
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sarthak20th:
thanks
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