Question

A particle is projected at an angle of 30° with ground with speed 40 m/s. The speed of the particle after 2 s is (use g=10 ms-2)

Answer 1

Initial velocity $(u) = 40 \;m/s$
Angle of projection $(θ) = 30°$
Acceleration due to gravity $(g) = 10 \;m/s²$

Horizontal component $(u_x) = u$ $\times$ $cos(θ)$
Vertical component $(u_y) = u \times sin(θ)$

Horizontal component $(u_x) = 40 \times cos(30°) ≈ 34.64 \;m/s$
Vertical component $(u_y) = 40 \times sin(30°) ≈ 20 \;m/s$

velocity after $2$ seconds: $v_y = u_y + gt$

Substituting the known values: $v_y = 20 + (10 \times 2) = 20 + 20 = 40 \;m/s$

Now, to find the resultant velocity after 2 seconds, we can use Pythagoras' theorem:

$v = \sqrt{(v_x² + v_y²)}$

$v = \sqrt{((34.64)²} + (40)²) v ≈ \sqrt{(1199.13 + 1600) }≈ \sqrt{(2799.13)} ≈ 52.92 \;m/s$

So, the speed of the particle after $2$ seconds is approximately $52.92$ $m/s$.