The equation of the path of the projectile is y=0.5x-0.04x^2.the initial speed of projectile is
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Equation of path of projectile is given by y = 0.5x - 0.04x²
or, y = x/2 - 10x²/(1000/4)
We know, standard equation of projectile is given by ,
y = xtanФ - gx²/2u²cos²Ф
Here, Ф is angle of projectile with horizontal
u is the intial speed of projectile and g is Acceleration due to gravity.
now, compare both equations,
tanФ = 1/2 , so, cosФ = 2/√5 -----(1)
g/2u²cos²Ф = 10/(1000/4)
2u²cos²Ф = 1000/4 = 250
2u² × 4/5 = 250 [ from equation (1),
u² = 625/4
so, u = 25/2 = 12.5 m/s
Hence, intial speed = 12.5 m/s
or, y = x/2 - 10x²/(1000/4)
We know, standard equation of projectile is given by ,
y = xtanФ - gx²/2u²cos²Ф
Here, Ф is angle of projectile with horizontal
u is the intial speed of projectile and g is Acceleration due to gravity.
now, compare both equations,
tanФ = 1/2 , so, cosФ = 2/√5 -----(1)
g/2u²cos²Ф = 10/(1000/4)
2u²cos²Ф = 1000/4 = 250
2u² × 4/5 = 250 [ from equation (1),
u² = 625/4
so, u = 25/2 = 12.5 m/s
Hence, intial speed = 12.5 m/s
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