1. Resolve the initial velocity into vertical component

Given: $u = 25\ \text{m/s}, \quad \theta = 45^\circ, \quad g = 10\ \text{m/s}^2$

The vertical component is $u_y = u\sin\theta = 25\sin45^\circ = 25\cdot \frac{1}{\sqrt{2}} = \frac{25}{\sqrt{2}}\ \text{m/s}$

2. Time to reach the highest point

At the highest point, vertical velocity becomes zero. Using: $v_y = u_y - gt$

So, $0 = \frac{25}{\sqrt{2}} - 10t$ $t = \frac{25}{10\sqrt{2}} = \frac{2.5}{\sqrt{2}} \approx 1.77\ \text{s}$

3. Maximum height

Use the formula: $H = \frac{u_y^2}{2g}$

Substitute $u_y = \frac{25}{\sqrt{2}}$: $H = \frac{\left(\frac{25}{\sqrt{2}}\right)^2}{2\cdot 10}$ $H = \frac{625/2}{20} = \frac{625}{40} = 15.625\ \text{m}$

4. Match with the options

We get:

• Maximum height $= 15.625\ \text{m}$
• Time to highest point $= 1.77\ \text{s}$

This matches Option C.