Given data

• Initial position: origin, so $x_0 = 0, \quad y_0^{(initial)} = 0$
• Initial velocity: $\vec u = 5\hat j \Rightarrow u_x = 0,\; u_y = 5\,\text{m/s}$
• Constant acceleration: $\vec a = 10\hat i + 4\hat j \Rightarrow a_x = 10,\; a_y = 4\,\text{m/s}^2$

Find time using x-coordinate

The particle is at $x=20$ m at time $t$.

Using $x = x_0 + u_x t + \frac{1}{2} a_x t^2$ we get $20 = 0 + 0\cdot t + \frac{1}{2}(10)t^2$ $20 = 5t^2$ $t^2 = 4$ $t = 2\,\text{s}$

(Taking positive time only.)

Find y-coordinate at this time

Using $y = y_0^{(initial)} + u_y t + \frac{1}{2} a_y t^2$ $y = 0 + 5(2) + \frac{1}{2}(4)(2^2)$ $y = 10 + 2\cdot 4$ $y = 10 + 8 = 18\,\text{m}$

So, $y_0 = 18\,\text{m}$

Match with options

• A: $5$ s and $25$ m ❌
• B: $2$ s and $18$ m ✅
• C: $2$ s and $24$ m ❌
• D: $4$ s and $52$ m ❌

Final answer

The correct option is B: $t=2\,\text{s},\quad y_0=18\,\text{m}$

Comparison with stored correct answer

Stored correct answer: B

My derived answer also gives B, so they agree.