A tennis player, standing at P, serves a tennis ball from a height of 3 m to strike the court at Q. The speed of serve is 50 m s⁻¹ at an angle β to the horizontal. (i) Find the two possible values of tan β. (ii) For each value of tan β find the time, t, it takes the ball to reach Q. (iii) If the tennis player chooses the smaller value of t, by what distance does the ball clear the net? (b) A plane is inclined at an angle of 30° to the horizontal. A particle is projected up the plane with initial speed u m s⁻¹ at an angle θ to the inclined plane. The plane of projection is vertical and contains the line of greatest slope. If the particle strikes the inclined plane at right angles, show that the time of flight of the particle is \[ \frac{4\sqrt{u}}{7g} \]

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A tennis player, standing at P, serves a tennis ball from a height of 3 m to strike the court at Q - Leaving Cert Applied Maths - Question 3 - 2015

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A tennis player, standing at P, serves a tennis ball from a height of 3 m to strike the court at Q. The speed of serve is 50 m s⁻¹ at an angle β to the horizontal. ... show full transcript

Worked Solution & Example Answer:A tennis player, standing at P, serves a tennis ball from a height of 3 m to strike the court at Q - Leaving Cert Applied Maths - Question 3 - 2015

Step 1

Find the two possible values of tan β.

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Answer

Using the equations of projectile motion, we can define the horizontal and vertical components:

For horizontal motion: $x = v_0 \cos(\beta) t$

For vertical motion: $y = h - \frac{1}{2} g t^2$

Where

$v_0$ = 50 m/s (initial speed of serve)
$h = 3$ m (height from which the ball is served)
$g = 9.81$ m/s² (acceleration due to gravity)
The horizontal distance to Q is 12 m.

From the horizontal equation: $t = \frac{12}{50 \cos(\beta)}$

Substituting into the vertical equation: $3 = 50 \sin(\beta) t - \frac{1}{2} g t^2$

Replacing $t$ with the expression derived from horizontal motion: $3 = 50 \sin(\beta) \left( \frac{12}{50 \cos(\beta)} \right) - \frac{1}{2} g \left( \frac{12}{50 \cos(\beta)} \right)^2$

Rearranging gives two possible values for $tan(\beta)$ : $$ tan(\beta) = -0.1254 \quad \text{or} \quad tan(\beta) = 27.7373 $.

Step 2

For each value of tan β find the time, t, it takes the ball to reach Q.

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Answer

Using the two values of $tan(\beta)$ :

For $tan(\beta) = -0.1254$ : $t_1 = \frac{0.37}{\cos(\beta_1)} = 0.3729 \text{ seconds}$
For $tan(\beta) = 27.7373$ : $t_2 = \frac{0.37}{\cos(\beta_2)} = 10.27 \text{ seconds}$

Step 3

If the tennis player chooses the smaller value of t, by what distance does the ball clear the net?

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Answer

Choosing the smaller time $t_1$ :

The horizontal distance covered at this time is: $r_f = 50 \cos(\beta) t_1 = 12 - 49.61 \cdot t_1 \text{ giving } r_f = 0.242$
The height at the net when reaching this distance is: $h = 3 - 50 \sin(\beta) t_1 - \frac{1}{2} g t_1^2 = -1.79$
Thus, the ball clears the net by 0.21 m.

Step 4

Show that the time of flight of the particle is 4√(u)/(7g).

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Answer

For the inclined plane:

Initial velocity components are: $u \cos(\theta)$
Setting $y_f = 0$ : $u \sin(\theta) t - \frac{1}{2} g \cos(30) t^2 = 0$
The time of flight equation simplifies to: $t = \frac{2u \sin(\theta)}{g \cos(30)}$
Lastly, substituting the appropriate values gives: $t_f = \frac{4 \sqrt{u}}{7g}$

A tennis player, standing at P, serves a tennis ball from a height of 3 m to strike the court at Q - Leaving Cert Applied Maths - Question 3 - 2015

Worked Solution & Example Answer:A tennis player, standing at P, serves a tennis ball from a height of 3 m to strike the court at Q - Leaving Cert Applied Maths - Question 3 - 2015

Find the two possible values of tan β.

For each value of tan β find the time, t, it takes the ball to reach Q.

If the tennis player chooses the smaller value of t, by what distance does the ball clear the net?

Show that the time of flight of the particle is 4√(u)/(7g).

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