The relation between the horizontal displacement x (in metre) and the vertical displacement y (in metre) of a projectile is y=3 x-0.8 x^2 . The time of flight of the projectile is (Acceleration due to gravity =10 ~ms ⁻² ).

Step-by-step solution

For a projectile, y=3 x-0.8 x^2 ....(i) aligned & At x=R, y=0 & 0=3 R-0.8 R^2 & R(0.8 R-3)=0 & R= 3 0.8 = 15 4 m aligned Also, y = x - gn ^2 2 u _ x ^2 ...(ii) Comparing eq(i) with eqn (ii), we get g 2 u _ x ^2 =0.8 u _ x = 10 2 0.8 = 5 2 ~m / s Now, R=u_x T T= R u_x = 15 4 5 2 =1.5 ~s