Question

a ball is kicked at an angle of 40° above the horizontal with a speed of 25 m/s. a gust of wind comes in and pushes the ball horizontally a further 7 m/s in the same direction. what is the resulting speed of the ball?

Explanation:

Step1: Resolve the initial velocity components

The initial velocity of the ball $v_0 = 25$ m/s and the launch - angle $\theta=40^{\circ}$. The initial horizontal velocity $v_{0x}=v_0\cos\theta$ and the initial vertical velocity $v_{0y}=v_0\sin\theta$.
$v_{0x}=25\cos40^{\circ}\approx25\times0.766 = 19.15$ m/s, $v_{0y}=25\sin40^{\circ}\approx25\times0.643 = 16.075$ m/s.
The wind pushes the ball horizontally with an additional velocity $v_{wind}=7$ m/s.

Step2: Calculate the new horizontal velocity

The new horizontal velocity $v_x=v_{0x}+v_{wind}=19.15 + 7=26.15$ m/s. The vertical velocity remains unchanged (assuming no vertical - wind effect), $v_y = v_{0y}\approx16.075$ m/s.

Step3: Calculate the resulting speed

The resulting speed $v$ of the ball is given by the Pythagorean theorem $v=\sqrt{v_x^{2}+v_y^{2}}$.
$v=\sqrt{(26.15)^{2}+(16.075)^{2}}=\sqrt{683.8225 + 258.405625}=\sqrt{942.228125}\approx30.7$ m/s.

Answer:

Approximately 30.7 m/s