Mark throws a ball with initial speed of 125 feet per second at an angle of 40 degrees. It was thrown 3 feet off the ground. How long was the ball in the air? How far did the ball travel horizontally? What was the maximum height of the ball?

use the parametric equations: x = (Vo cos theta)t , y = h + (Vo sin theta)t-16t^2

Question

29-05-2023
Mathematics

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Mark throws a ball with initial speed of 125 feet per second at an angle of 40 degrees. It was thrown 3 feet off the ground. How long was the ball in the air? How far did the ball travel horizontally? What was the maximum height of the ball?

use the parametric equations: x = (Vo cos theta)t , y = h + (Vo sin theta)t-16t^2

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semsee45 semsee45 · Answer 1 · 2023-05-29T11:42:24+02:00

Answer:

The ball was in the air for 5.06 seconds (2 d.p.).

The ball travelled 484.41 feet (2 d.p.) horizontally.

The maximum height of the ball is 103.87 feet (2 d.p.).

Step-by-step explanation:

When a body is projected through the air with initial speed (v₀), at an angle of θ to the horizontal, it will move along a curved path.

Therefore, trigonometry can be used to resolve the body's initial velocity into its vertical and horizontal components.

If a ball is thrown at an initial velocity (v₀) of 125 ft/s at an angle of 40°, then:

Horizontal component of v₀ = 125 cos 40°
Vertical component of v₀ = 125 sin 40°

The given parametric equations model the horizontal and vertical distances of the ball.

Substitute v₀ = 125 and θ = 40° into the given equations.

As the ball was thrown 3 ft off the ground, substitute h = 3.

Therefore, the equations that model the horizontal and vertical distances of the ball are:

[tex]x=(125 \cos 40^{\circ})t[/tex]

[tex]y=3+(125 \sin40^{\circ})t-16t^2[/tex]

The ball will stop travelling when its vertical distance from the ground is zero, i.e. y = 0.

Set the parametric equation for y to zero and solve for t:

[tex]\begin{aligned} \implies 0&=3+(125 \sin 40^{\circ})t-16t^2\\0&=-16t^2+(125 \sin 40^{\circ})t+3\\\\\implies t&=5.05884201...\; \sf s\\t&= -0.0370638...\; \sf s\end{aligned}[/tex]

As time is positive only, the ball was in the air for 5.06 seconds (2 d.p.).

To find the distance the ball travelled horizontally, substitute the found value of t into the parametric equation for x:

[tex]x=(125 \cos 40^{\circ})t[/tex]

[tex]x=(95.7555553...) (5.05884201...)[/tex]

[tex]x=484.41222...[/tex]

[tex]x=484.41\; \sf ft\;(2\;d.p.)[/tex]

Therefore, the ball travelled 484.41 feet horizontally.

When the ball reaches its maximum height, the vertical component of its velocity is momentarily zero.

To find the time when the vertical component of its velocity is zero, we can use the kinematic formula:

[tex]\boxed{v = v_0 + at}[/tex]

where:

v is velocity (in ft s⁻¹).
v₀ is initial velocity (in ft s⁻¹).
a is acceleration due to gravity (32 ft s⁻²).
t is time (in seconds).

Therefore, taking ↑ as positive:

v = 0
v₀ = 125 sin 40°
a = -32

Substitute these values into the formula and solve for t:

[tex]\begin{aligned}v&=v_0+at\\\implies 0&=125 \sin 40^{\circ}-32t\\32t&=125 \sin 40^{\circ}\\t&=\dfrac{125 \sin 40^{\circ}}{32}\\t&=2.5108891\; \sf s\end{aligned}[/tex]

Therefore, the ball was at its maximum height at 2.51 s.

To find the maximum height, substitute the found value of t into the equation for y:

[tex]y=3+(125 \sin40^{\circ})(2.5108891)-16(2.5108891)^2[/tex]

[tex]y=103.873025...[/tex]

[tex]y=103.87\; \sf ft\;(2\;d.p.)[/tex]

Therefore, the maximum height of the ball is 103.87 feet (2 d.p.).