A particle is projected at an angle of $$30^{o}$$ from horizontal at a speed of $$ 60 m/s.$$ The height traversed by the particle in the first second is $$ h_0 $$ and height traversed in the last second, before it reaches the maximum height, is $$ h_1.$$ The ratio $$h_0:h_1 $$ is _______[Take, $$g=10m/s^{2}]$$

A particle is projected at $$30°$$ from horizontal at 60 m/s. We need to find the ratio $$h_0 : h_1$$.

$$u_y = u \sin 30° = 60 \times \frac{1}{2} = 30$$ m/s

At maximum height, $$v_y = 0$$:

$$0 = u_y - gt \Rightarrow t = \frac{u_y}{g} = \frac{30}{10} = 3$$ s

$$h_0 = u_y(1) - \frac{1}{2}g(1)^2 = 30 - 5 = 25$$ m

The last second is from t = 2 to t = 3.

Height at t = 2: $$y_2 = 30(2) - \frac{1}{2}(10)(4) = 60 - 20 = 40$$ m

Height at t = 3: $$y_3 = 30(3) - \frac{1}{2}(10)(9) = 90 - 45 = 45$$ m

$$h_1 = y_3 - y_2 = 45 - 40 = 5$$ m

$$h_0 : h_1 = 25 : 5 = 5 : 1$$

The answer is 5.