What is projectile motion and its deivation explain
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Let us say the the project is sent in to air at an angle of Ф with the horizontal and with an initial speed of u in that direction.
There is a deceleration in the vertical direction = -g.
The speed of projectile in the vertical direction is : v = u + at =>
Vy = u sin Ф - g t -- equation 1
The speed in the horizontal direction is : Vx = u cos Ф. -- equation 2
There is no acceleration in X-direction.
The distance travelled is : s = ut+1/2 at²
=> Sx = x = u (cos Ф) t --- equation 3
t = x / (u cos Ф)
Sy = y = (u Sin Ф) t - 1/2 g t²
y = u sinФ x / u cosФ - 1/2 g x² / u² Cos² Ф
y = x tan Ф - g x² / (2u²Cos²Ф) --- equation 4
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y = H when Vy becomes 0. => Vy = 0 = u sin Ф - g t
=> t = u sin Ф / g
y = u Sin Ф * (u sin Ф / g) - 1/2 g (u SinФ/g)²
y = H = u² Sin² Ф / 2 g -- equation 5
y = 0 => x = R
Substituting in equation 4, we get R = u² Sin 2Ф / g -- equation 6
There is a deceleration in the vertical direction = -g.
The speed of projectile in the vertical direction is : v = u + at =>
Vy = u sin Ф - g t -- equation 1
The speed in the horizontal direction is : Vx = u cos Ф. -- equation 2
There is no acceleration in X-direction.
The distance travelled is : s = ut+1/2 at²
=> Sx = x = u (cos Ф) t --- equation 3
t = x / (u cos Ф)
Sy = y = (u Sin Ф) t - 1/2 g t²
y = u sinФ x / u cosФ - 1/2 g x² / u² Cos² Ф
y = x tan Ф - g x² / (2u²Cos²Ф) --- equation 4
====================================
y = H when Vy becomes 0. => Vy = 0 = u sin Ф - g t
=> t = u sin Ф / g
y = u Sin Ф * (u sin Ф / g) - 1/2 g (u SinФ/g)²
y = H = u² Sin² Ф / 2 g -- equation 5
y = 0 => x = R
Substituting in equation 4, we get R = u² Sin 2Ф / g -- equation 6
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TPS:
Nice answer sir!!!
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