A particle is projected at an angle of ${30}^{\circ }$ from horizontal at a speed of$60\mathrm{m}/\mathrm{s}$. 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______
$[\text{Take,}g=10\mathrm{m}/{\mathrm{s}}^2]$

Given Data:

Initial velocity: $u = 60$ m/s

Angle of projection: $\theta = 30^\circ$

Acceleration due to gravity: $g = 10$ m/s²

We need to find the ratio of:

Height traversed in the first second ($h_0$)

Height traversed in the last second before reaching maximum height ($h_1$)

Step 1: Find the Vertical Component of Velocity

The initial vertical velocity:

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

Step 2: Find the Time to Reach Maximum Height

At the maximum height, the vertical velocity becomes zero:

$$v_y = u_y - g t$$

 $$0 = 30 - 10t$$ 

$$t = \frac{30}{10} = 3 \text{ s}$$

So, the total time to reach maximum height is 3 seconds.

Step 3: Height Traversed in the First Second ($h_0$)

Using the equation of motion:

$h_0=u_{y}t+\frac{1}{2}{a}_y{t}^2$

For $t = 1$ s:

$h_0=(30\times 1)+\frac{1}{2}(-10\times 1^2)$

 $$h_0 = 30 - 5 = 25 \text{ m}$$ 

Step 4: Height Traversed in the Last Second ($h_1$)

The height traversed in the last second before reaching maximum height is given by:

$h_1=v_{y}t-\frac{1}{2}{a}_y{t}^2$

 ${\mathrm{h}}_1=0-\frac{1}{2}(-10\times 1^2)=5\mathrm{m}$  

Step 5: Find the Ratio $h_0 : h_1$

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