A particle is projected at an angle of 30° from horizontal at a speed of 60 m/s. The height traversed by the particle in the first second is h0 and height traversed in the last second, before it reaches the maximum height, is h1. The ratio h0 : h1 is ______ .

Key ideas you need

1. Vertical component of velocity
   $u_y = u\sin\theta$
   For $u = 60\;\text{m/s}$ and $\theta = 30\degree$, $\sin 30\degree = \tfrac12$, so
   $u_y = 60 \times \frac12 = 30\;\text{m/s}$

2. Equation of vertical displacement
   For an upward‐positive axis and constant gravity $g$ (take $g = 10\;\text{m/s}^2$ to keep numbers simple):
   $y(t) = u_y t - \frac12 g t^2$ This gives the vertical position at any time $t$ measured from the launch point.

3. Time to the highest point
   The ball stops rising when its vertical velocity becomes zero:
   $v_y = u_y - g t = 0 \;\Longrightarrow\; t_{\text{top}} = \frac{u_y}{g}$
   The duration from launch to that instant is crucial because the "last second" must end exactly at this time.

4. Height in a particular one–second interval
   Height climbed between $t = t_1$ and $t = t_2$ is simply
   $\Delta h = y(t_2) - y(t_1)$
   Identify the correct limits for the first second ($0\to1$) and the last second ( $t_{\text{top}}-1 \to t_{\text{top}}$ ).

Logical chain to attack the problem

1. Find $u_y$.
2. Compute $t_{\text{top}}$.
3. Plug into $y(t)$ to get positions at $t = 0, 1, (t_{\text{top}}-1), t_{\text{top}}$.
4. Subtract to obtain $h_0$ and $h_1$.
5. Form the ratio $\dfrac{h_0}{h_1}$.

Imagine throwing a ball into the air

1. Throwing speed and angle
   You throw a ball at a certain slanted angle, here $30\degree$, with a speed of $60\;\text{m/s}$.
2. Upward and sideways motion
   The throw splits into two parts: up–down and left–right. We only care about the up–down (vertical) part to find heights.
3. First second vs. last second
   • First second: how high did the ball climb from the very beginning up to $1\;\text{s}$?
   • Last second before the peak: when the ball is about to stop going up, the very last $1\;\text{s}$ chunk of that climb is tiny.
4. Why do they differ?
   Gravity keeps slowing the ball. At the start it is fastest upward, so it covers a lot of height. Near the top it is slow, so in the last second it rises only a little.

Step 1: Resolve initial velocity

$u_y = u\sin\theta = 60 \times \frac12 = 30\;\text{m/s}$

Step 2: Time to reach maximum height

$t_{\text{top}} = \frac{u_y}{g} = \frac{30}{10} = 3\;\text{s}$

Step 3: Write vertical position function

$y(t) = u_y t - \frac12 g t^2 = 30 t - 5 t^2$

Step 4: Height in the first second

Position at $t = 1\;\text{s}$:
$y(1) = 30(1) - 5(1)^2 = 30 - 5 = 25\;\text{m}$ Position at launch $t = 0$: $y(0)=0$.
Therefore
$h_0 = y(1) - y(0) = 25\;\text{m}$

Step 5: Height in the last second before the top

Last second runs from $t = 2\;\text{s}$ to $t = 3\;\text{s}$.

Position at $t = 3\;\text{s}$ (the peak):
$y(3) = 30(3) - 5(3)^2 = 90 - 45 = 45\;\text{m}$

Position at $t = 2\;\text{s}$:
$y(2) = 30(2) - 5(2)^2 = 60 - 20 = 40\;\text{m}$

Thus
$h_1 = y(3) - y(2) = 45 - 40 = 5\;\text{m}$

Step 6: Form the ratio

$\frac{h_0}{h_1} = \frac{25}{5} = 5:1$

[\boxed{,h_0 : h_1 = 5 : 1,}]