Body is thrown which covers a maximum height of 50m and range 25m find angle of projection????
Asked by riyaz1111 | 25th Jan, 2019, 05:22: PM
Let 'u' be the velocity of projection and (Θ) be the angle of projection , then
Range of projectile R,
R= u2 sin (2Θ)/g…………(1).
here, 'g' is the acceleration due to gravity.
We have maximum height H,
H= u2 sin2 (Θ)/(2g)…………(2).
Dividing equation (2) by equation (1)
H/R= tan(Θ)/4
Substituting for H and R,
50/25= tan(Θ)/4
Tan (Θ) =8
Θ = tan⁻¹ (8)
Θ = 82.87o
Angle of projection is 82.87o
Let 'u' be the velocity of projection and (Θ) be the angle of projection , then
Range of projectile R,
R= u2 sin (2Θ)/g…………(1).
here, 'g' is the acceleration due to gravity.
We have maximum height H,
H= u2 sin2 (Θ)/(2g)…………(2).
Dividing equation (2) by equation (1)
H/R= tan(Θ)/4
Substituting for H and R,
50/25= tan(Θ)/4
Tan (Θ) =8
Θ = tan⁻¹ (8)
Θ = 82.87o
Angle of projection is 82.87o
Answered by Science Mate | 25th Jan, 2019, 05:36: PM
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