The parametric equations are the equations for the horizontal and
vertical motions.
Responses:
a.) x = 25·t, y = 25·t·√3 - 16·t²
b.) 2.8 s
c.) 70 ft.
The given parameters are;
The angle at which the tennis ball leaves the racket, θ  = 60°
The initial speed of the tennis ball, u = 50 ft./s
The initial height of the tennis ball, yâ‚€ = 5 ft.
a.) The vertical y motion of the ball is under gravitational acceleration,
while the horizontal x motion is constant.
Therefore;
The horizontal velocity = 50·cos(60°) = 50 × 0.5 = 25
Therefore;
The vertical distance is given by the equation;
y = y₀ + [tex]\mathbf{u_y}[/tex]·t - 0.5·32·t²
Where;
[tex]u_y = \mathbf{u \cdot sin(\theta)}[/tex]
Which gives;
[tex]u_y = 50 \times sin \left(60 ^{\circ}\right) = 50 \times \dfrac{\sqrt{3} }{2} = \mathbf{25 \cdot \sqrt{3}}[/tex]
Therefore;
The parametric equations are;
b.) The duration the ball will be in the air is given by the equation;
y = 0 = 5 + 25·t·√3 - 16·t²
Which gives;
[tex]t = \mathbf{\dfrac{-25 \cdot \sqrt{3} \pm\sqrt{\left(25 \cdot \sqrt{3} \right)^2 - 4 \times (-16) \times 5 } }{2 \times (-16)}}[/tex]
t ≈ -0.111 s, or t ≈ 2.8 s
c.) The distance the tennis ball travelled in the air are;
The time at which the ball is at maximum height, we have;
[tex]t = -\dfrac{25 \cdot \sqrt{3} }{2 \times (-16)} = \dfrac{25 \cdot \sqrt{3} }{32}[/tex]
Which gives;
[tex]y_{max} = 5 + \dfrac{25 \cdot \sqrt{3} }{32} \times 25 \cdot \sqrt{3} - 16 \times \left( \dfrac{25 \cdot \sqrt{3} }{32} \right)^2 \approx 34.3[/tex]
The maximum height, [tex]y_{max}[/tex] ≈ 34.3 ft.
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