A projectile is projected at 30∘ from horizontal with initial velocity 40 ms−1. The velocity of the projectile at t=2 s from the start will be: (Given g=10 m/s2 )

To solve this problem, we need to find the horizontal and vertical components of the velocity at t=2s.

Step 1: Find the initial horizontal and vertical components of the velocity. The initial velocity (u) is given as 40 m/s and the angle (θ) is 30 degrees. We can use the equations:

u_x = ucos(θ) for the horizontal component and u_y = usin(θ) for the vertical component.

Using these equations, we get:

u_x = 40cos(30) = 34.64 m/s (approx) and u_y = 40sin(30) = 20 m/s.

Step 2: Find the vertical velocity at t=2s. The vertical velocity (v_y) at any time t can be found using the equation:

v_y = u_y - g*t

where g is the acceleration due to gravity. Substituting the given values, we get:

v_y = 20 - 10*2 = 0 m/s.

Step 3: Find the resultant velocity at t=2s. The horizontal component of the velocity (u_x) remains constant throughout the motion. So, at t=2s, the horizontal velocity is still 34.64 m/s.

The resultant velocity (v) can be found using the equation:

v = sqrt((u_x)^2 + (v_y)^2)

Substituting the values, we get:

v = sqrt((34.64)^2 + 0) = 34.64 m/s.

So, the velocity of the projectile at t=2s from the start is approximately 34.64 m/s.