Chapter 6.1, Problem 45E

Solution

(a)

To find: The component form of the initial velocity.

The component form of the initial velocity is equal to $〈3.42,9.40〉$ .

Given information:

A basketball is shot at a ${70}^{\circ }$ angle with the horizontal direction with an initial speed of $10$ m/sec.

Concept used:

The component form of a vector is $\overrightarrow{a}=〈\left|a\right|\cos \theta ,\left|a\right|\sin \theta 〉$ .

Calculation:

Apply the above concept and find the component form of the initial velocity vector $\overrightarrow{v}$ below:

  $\begin{array}{c}\overrightarrow{v}=〈10\cos {70}^{\circ },10\sin {70}^{\circ }〉\\ =〈3.42,9.40〉\end{array}$

(b)

To find: The horizontal component and vertical component of the velocity.

The horizontal component is equal to $3.42$ and the vertical component is equal to $9.40$ .

Given information:

A basketball is shot at a ${70}^{\circ }$ angle with the horizontal direction with an initial speed of $10$ m/sec.

Concept used:

The horizontal component of a vector is $\left|a\right|\cos \theta$ .

The vertical component of a vector is $\left|a\right|\sin \theta$ .

Calculation:

Apply the above concept and find the horizontal component and vertical component below:

The horizontal component of the velocity is $10\cos {70}^{\circ }=3.42$ .

The vertical component of the velocity is $10\sin {70}^{\circ }=9.40$ .