Chapter 6.3, Problem 50E

Solution

a)

To find: the distance traveled by the golf ball if the initial velocity is $15°$ .

The distance traveled by a golf ball is $506.25$ ft.

Given information:

The initial velocity of the golf ball is 180 ft//sec

The angle of elevation of the ball is $15°$

Formula used:

• Suppose that an object is thrown from a point $y_0$ above the ground level with an initial speed of $v_0$ ft/sec at an angle $\theta$ with the horizontal. The path of the object is modeled by parametric equations $x=\left(v_0\cos \theta \right)t,$ $y=-16t^2+\left(v_0\sin \theta \right)t+y_0$ .

Calculation:

To get the golf ball’s path, substitute $180$ for $v_0$ , $15°$ for $\theta$ , and 0 for $y_0$ in parametric equations: $x=\left(v_0\cos \theta \right)t,$

  $y=-16t^2+\left(v_0\sin \theta \right)t+y_0$ .

  $x=\left(180\cos 15°\right)t,$

  $y=-16t^2+\left(180\sin 15°\right)t$

At $t=0$ , the initial point $\left(x,y\right)=\left(0,0\right)$ .

The golf ball hit the ground at the point $\left(\left(180\cos 15°\right)t,0\right)$ for some $t$ .

To find the value $t$ , solve the equation $-16t^2+\left(180\sin 15°\right)t=0$ for $t$ .

  $\begin{array}{l}-16t^2+\left(180\sin 15°\right)t=0\\ -t\left(16t-46.59\right)=0\\ t=0,2.91\end{array}$

Thus, the golf ball hit the ground after 2.91 seconds. The distance travelled by golf ball is

  $\begin{array}{c}x=\left(180\cos 15°\right)t\\ =\left(173.97\right)\left(2.91\right)\\ =506.25\end{array}$

Thus, the golf ball hit the ground at $t=2.91$ sec. and distance traveled is $506.25$ ft.

b)

To find: the distance traveled by golf ball if the initial velocity is $20°$ .

The distance traveled by golf ball is $650.80$ ft.

Given information:

The initial velocity of the golf ball is 180 ft//sec

The angle of elevation of the ball is $20°$

Formula used:

• Suppose that an object is thrown from a point $y_0$ above the ground level with an initial speed of $v_0$ ft/sec at an angle $\theta$ with the horizontal. The path of the object is modeled by parametric equations $x=\left(v_0\cos \theta \right)t,$ $y=-16t^2+\left(v_0\sin \theta \right)t+y_0$ .

Calculation:

To get the golf ball’s path, substitute $180$ for $v_0$ , $20°$ for $\theta$ , and 0 for $y_0$ in parametric equations: $x=\left(v_0\cos \theta \right)t,$

  $y=-16t^2+\left(v_0\sin \theta \right)t+y_0$ .

  $x=\left(180\cos 20°\right)t,$

  $y=-16t^2+\left(180\sin 20°\right)t$

At $t=0$ , the initial point $\left(x,y\right)=\left(0,0\right)$ .

The golf ball hit the ground at the point $\left(\left(180\cos 20°\right)t,0\right)$ for some $t$ .

To find the value $t$ , solve the equation $-16t^2+\left(180\sin 20°\right)t=0$ for $t$ .

  $\begin{array}{l}-16t^2+\left(180\sin 20°\right)t=0\\ -t\left(16t-61.56\right)=0\\ t=0,3.85\end{array}$

Thus, the golf ball hit the ground after 3.85 seconds. The distance travelled by golf ball is

  $\begin{array}{c}x=\left(180\cos 20°\right)t\\ =\left(169.14\right)\left(3.85\right)\\ =650.80\end{array}$

Thus, the golf ball hit the ground at $t=3.85$ sec. and traveled distance is $650.80$ ft.

c)

To find: the distance traveled by golf ball if the initial velocity is $25°$ .

The distance traveled by golf ball is $775.61$ ft.

Given information:

The initial velocity of the golf ball is 180 ft//sec

The angle of elevation of the ball is $25°$

Formula used:

• Suppose that an object is thrown from a point $y_0$ above the ground level with an initial speed of $v_0$ ft/sec at an angle $\theta$ with the horizontal. The path of the object is modeled by parametric equations $x=\left(v_0\cos \theta \right)t,$ $y=-16t^2+\left(v_0\sin \theta \right)t+y_0$ .

Calculation:

To get the golf ball’s path, substitute $180$ for $v_0$ , $25°$ for $\theta$ , and 0 for $y_0$ in parametric equations: $x=\left(v_0\cos \theta \right)t,$

  $y=-16t^2+\left(v_0\sin \theta \right)t+y_0$ .

  $x=\left(180\cos 25°\right)t,$

  $y=-16t^2+\left(180\sin 25°\right)t$

At $t=0$ , the initial point $\left(x,y\right)=\left(0,0\right)$ .

The golf ball hit the ground at the point $\left(\left(180\cos 25°\right)t,0\right)$ for some $t$ .

To find the value $t$ , solve the equation $-16t^2+\left(180\sin 25°\right)t=0$ for $t$ .

  $\begin{array}{l}-16t^2+\left(180\sin 25°\right)t=0\\ -t\left(16t-76.0712\right)=0\\ t=0,4.7544\end{array}$

Thus, the golf ball hit the ground after 4.75 seconds. The distance travelled by golf ball is

  $\begin{array}{c}x=\left(180\cos 25°\right)t\\ =\left(163.1354\right)\left(4.7544\right)\\ =775.61\end{array}$

Thus, the golf ball hit the ground at $t=4.75$ sec. and distance traveled is $775.61$ ft.

c)

To find: the distance traveled by golf ball if its initial velocity is $30°$ .

The distance traveled by golf ball is $876.85$ ft.

Given information:

The initial velocity of the golf ball is 180 ft//sec

The angle of elevation of the ball is $30°$

Formula used:

• Suppose that an object is thrown from a point $y_0$ above the ground level with an initial speed of $v_0$ ft/sec at an angle $\theta$ with the horizontal. The path of the object is modeled by parametric equations $x=\left(v_0\cos \theta \right)t,$ $y=-16t^2+\left(v_0\sin \theta \right)t+y_0$ .

Calculation:

To get the golf ball’s path, substitute $180$ for $v_0$ , $30°$ for $\theta$ , and 0 for $y_0$ in parametric equations: $x=\left(v_0\cos \theta \right)t,$

  $y=-16t^2+\left(v_0\sin \theta \right)t+y_0$ .

  $x=\left(180\cos 30°\right)t,$

  $y=-16t^2+\left(180\sin 30°\right)t$

At $t=0$ , the initial point $\left(x,y\right)=\left(0,0\right)$ .

The golf ball hit the ground at the point $\left(\left(180\cos 30°\right)t,0\right)$ for some $t$ .

To find the value $t$ , solve the equation $-16t^2+\left(180\sin 30°\right)t=0$ for $t$ .

  $\begin{array}{l}-16t^2+\left(180\sin 30°\right)t=0\\ -t\left(16t-90\right)=0\\ t=0,5.625\end{array}$

Thus, the golf ball hit the ground after 5.625 seconds. The distance travelled by golf ball is

  $\begin{array}{c}x=\left(180\cos 30°\right)t\\ =\left(155.8846\right)\left(5.625\right)\\ =876.85\end{array}$

Thus, the golf ball hit the ground at $t=5.625$ sec and reached the distance is $876.85$ ft.