A projectile is fired at an angle of 30° to the horizontal with a velocity of 40 m/s Calculate the velocity attained after 1 s. [g = 10 m/s\(^2\)]

Answer Details

Resolve the initial velocity. With \(u = 40\ \text{m/s}\) at \(30^{\circ}\) to the horizontal:

\[ u_x = 40\cos30^{\circ} = 40\times0.866 = 34.6\ \text{m/s} \] \[ u_y = 40\sin30^{\circ} = 40\times0.5 = 20\ \text{m/s} \]

After t = 1 s. The horizontal component is unchanged (no horizontal force):

\[ v_x = 34.6\ \text{m/s} \]

The vertical component is reduced by gravity:

\[ v_y = u_y - g t = 20 - 10\times1 = 10\ \text{m/s} \]

Resultant velocity.

\[ v = \sqrt{v_x^{2}+v_y^{2}} = \sqrt{34.6^{2}+10^{2}} = \sqrt{1197+100} = \sqrt{1297} \approx 36.0\ \text{m/s} \]

Direction above the horizontal.

\[ \tan\theta = \frac{v_y}{v_x} = \frac{10}{34.6} = 0.289 \;\Rightarrow\; \theta \approx 16.1^{\circ} \]

The velocity after 1 s is about 36.0 m/s directed at about 16° above the horizontal.