A body is projected at an angle of 30 with the horizontal and with a speed of 30 m/s. What is the angle with the horizontal after 1.5 s.
Answers
Answered by
2
For a projectile with an initial velocity u along an angle of projection Ф, the velocities along x and y at time t are :
v_x = u cosФ
v_y = u sin Ф - g t
Angle of the resultant velocity v with x axis:
= Tan⁻¹ (v_y / v_x)
= Tan⁻¹ [ (u sinФ - g t) / (u cosФ) ]
= Tan⁻¹ [ (30/2 - 10 *1.5) / (30 √3/2)
= 0 deg.
So the projectile is at the highest point of path.
v_x = u cosФ
v_y = u sin Ф - g t
Angle of the resultant velocity v with x axis:
= Tan⁻¹ (v_y / v_x)
= Tan⁻¹ [ (u sinФ - g t) / (u cosФ) ]
= Tan⁻¹ [ (30/2 - 10 *1.5) / (30 √3/2)
= 0 deg.
So the projectile is at the highest point of path.
Similar questions