Question

An athlete throws a shot at an angle of to the horizontal at an initial speed of 43 fts. It leaves his hand 7 ft above the ground.

(a) Where is the shot 2 seconds later?

(b) How high does the shot go?

(c) Where does the shot land? Exercise 55 in Section 10.

Short answer

(a)Therefore, 2 seconds, the height of the shot above the ground is 3.81 ft and the distance of the shot from the athlete is 60.81ft.

(b)The maximum height reached by the shot is, 14.45+7=21.45 ft.

(c) Therefore, the shot lands 64.16 ft away from the athelete.

Step-by-step solution

Concept of the motion

A projectile is an item on which gravitation is the only force that acts. Gravity affects the projectile's vertical motion, leading in vertical acceleration. The projectile's horizontal movement is caused by the tendency for any traveling item to maintain a constant velocity.

 vertical motion

(a)

The initial vertical velocity of the shot is \({v_0}\sin \theta = 43\sin {45^ \circ } = \frac{{43}}{{\sqrt 2 }}\)

The acceleration is a constant =\( - g = - 32ft/{s^2}\),

Use the following formula to find the vertical displacement of the shot after time, t=2,

Substitude\(u = \frac{{43}}{{\sqrt 2 }},t = 2\)and a= -32,

\(\begin{aligned}{*{20}{c}}{s = \frac{{43}}{{\sqrt 2 }} \times 2 - \frac{1}{2} \times 32 \times {{(2)}^2}}\\{s = 43\sqrt 2 - 64 \approx - 3.19}\end{aligned}\)

The shot was thrown from the height of 7 ft, and the displacement after 2 seconds is -3.19 ft.

Therefore, the height of the shot after 2 seconds is 7-3.19 =3.81ft.

Horizental motion

The initial horizontal velocity of the shot is,\({v_0}\cos \theta = 43\cos {45^ \circ } = \frac{{43}}{{\sqrt 2 }}\)

Since the acceleration is zero along with the horizontal direction, horizontal velocity remains a constant throughout the motion.

After two seconds the horizontal distance of the shot from the athlets is,

\(\frac{{43}}{{\sqrt 2 }} \times 2 = 43\sqrt 2 \approx 60.81ft\)

Therefore, 2 seconds, the height of the shot above the ground is 3.81 ft and the distance of the shot from the athlete is 60.81ft.

The height of the shot

(b)

The initial vertical velocity of the shot is \({v_0}\sin \theta = 43\sin {45^ \circ } = \frac{{43}}{{\sqrt 2 }}\)

Since the acceleration is constant = \( - g = - 32ft/{s^2}\),

When the shot reaches it’s maximum height, it’s vertical velocity is zero.

Use the following formula to find the maximum displacement,

Substitude,\(u = \frac{{43}}{{\sqrt 2 }},a = - 32\;and\;v = 0\)

\(\begin{aligned}{*{20}{c}}{2( - 32)s = {0^2} - {{\left( {\frac{{43}}{{\sqrt 2 }}} \right)}^2}}\\{s = \frac{1}{{64}} \cdot \frac{{1849}}{2} = \frac{{1849}}{{128}} \approx 14.45ft}\end{aligned}\)

The initial height of the shot was 7ft, and the maximum displacement of the shot is 14.45 ft.

The maximum height reached by the shot is, 14.45+7=21.45 ft.

Vertical motion

(c)

The initial vertical velocity of the shot is\({v_0}\sin \theta = 43\sin {45^ \circ } = \frac{{43}}{{\sqrt 2 }}\)

The acceleration is a constant =\( - g = - 32{\rm{ft}}/{{\rm{s}}^2}\).

Using the following formula,

Substitude,\(u = \frac{{43}}{{\sqrt 2 }},s = - 7\;and\;a = - 32\)

\(\begin{aligned}{*{20}{c}}{ - 7 = \frac{{43t}}{{\sqrt 2 }} - \frac{1}{2} \times 32 \times {t^2}}\\{0 = 7 + \frac{{43t}}{{\sqrt 2 }} - 16{t^2}}\\{16{t^2} - 30.41t - 7 = 0}\end{aligned}\)

Using the quardratic formula, we get

\(\begin{aligned}{*{20}{c}}{t = \frac{{ - ( - 30.41) \pm \sqrt {{{( - 30.41)}^2} - 4(16)( - 7)} }}{{2(16)}}}\\{t = \frac{{30.41 \pm \sqrt {924.77 + 448} }}{{32}}}\end{aligned}\)

\(t \approx - 0.21\;and\;t \approx 2.11\)

t cannot be negative, therefore we reject the negative solution.

Therefore, the ball hits the ground after 2.11 seconds.

Horizental motion

The intial horizontal velocity of the shot is\({v_0}\cos \theta = 43\cos {45^ \circ } = \frac{{43}}{{\sqrt 2 }}\)

Since the acceleration is zero along the horizontal direction, horizontal velocity remains a constant throughout the motion.

After 2.11 seconds, the horizontal distance of the shot from the athelete is

\(\frac{{43}}{{\sqrt 2 }} \times 2.11 \approx 64.16ft\)

Therefore, the shot lands 64.16 ft away from the athelete.