Answer: 3.5 m above its point of projection at approximately 2.94 seconds after being projected vertically upwards.
Explanation: The initial speed of the particle is given as 40 m/s, and the acceleration due to gravity is 10 m/s^2 (g = 10 m/s^2). We can use the equations of motion to solve this problem.
Let's assume the point of projection as the reference point, where the particle is at a height of 0 m. We want to find the times when the particle reaches a height of 3.5 m.
We can use the equation of motion: h = u*t - 0.5*g*t^2, where h is the height, u is the initial velocity, g is the acceleration due to gravity, and t is time.
Substituting the given values, we have: 3.5 m = 40 m/s * t - 0.5 * 10 m/s^2 * t^2.
Simplifying the equation, we get: 0.5 * 10 m/s^2 * t^2 - 40 m/s * t + 3.5 m = 0.
This equation is a quadratic equation in terms of t. We can solve it using the quadratic formula: t = (-b ± √(b^2 - 4ac)) / 2a, where a, b, and c are the coefficients of the equation.
Plugging in the values, we get: t = (-(-40) ± √((-40)^2 - 4 * 0.5 * 10 * 3.5)) / (2 * 0.5 * 10).
Simplifying further, we find: t = (40 ± √(1600 - 70)) / 10.
Calculating inside the square root, we get: t = (40 ± √(1530)) / 10.
Since we want to find the times when the particle is at a height of 3.5 m, we only consider the positive root of the equation.
Evaluating the positive root, we have: t = (40 + √(1530)) / 10.
Calculating this, we find: t ≈ 2.94 seconds.
Therefore, the particle will be 3.5 m above its point of projection at approximately 2.94 seconds after being projected vertically upwards.