Question

a volleyball was hit into the air at a speed of 58 miles per hour at an angle of 45° from the horizontal. express this velocity in vector form. round your answer to four decimals.

Explanation:

Step1: Find the horizontal component

The horizontal component \(v_x\) of a vector with magnitude \(v\) and angle \(\theta\) is given by \(v_x = v\cos\theta\). Here, \(v = 58\) and \(\theta=45^{\circ}\), so \(v_x=58\cos45^{\circ}\). Since \(\cos45^{\circ}=\frac{\sqrt{2}}{2}\approx0.7071\), then \(v_x = 58\times0.7071 = 41.0118\).

Step2: Find the vertical component

The vertical component \(v_y\) of a vector with magnitude \(v\) and angle \(\theta\) is given by \(v_y = v\sin\theta\). Here, \(v = 58\) and \(\theta = 45^{\circ}\), so \(v_y=58\sin45^{\circ}\). Since \(\sin45^{\circ}=\frac{\sqrt{2}}{2}\approx0.7071\), then \(v_y=58\times0.7071 = 41.0118\).

In vector form, the velocity vector \(\vec{v}=\langle v_x,v_y
angle\)

Answer:

\(\langle41.0118,41.0118
angle\)