A projectile is projected from origin having vertical as y-axis. The position of particle is given by x = 6t and y = 8t - 5t2 where x and y are in meter. If angle of projection in degree to nearest integer is (k + 50), then find k
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Given : A projectile is projected from origin having vertical as y-axis. The position of particle is given by x = 6t and y = 8t - 5t² where x and y are in meter.
angle of projection in degree to nearest integer is (k + 50),
To find : k
Solution:
y = 8t - 5t²
dy/dt = 8 - 10t
x = 6t
dx/dt = 6
dy/dx = (8 - 10t)/6 = (4 - 5t)/3
at t = 0 , x = 0 , y = 0
dy/dx = 4/3
tanα = 4/3
=> α = 53.13° ≈ 53°
Hence,
k + 50 = 53
=> k = 3
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