Drop shot 1

Relevant wiki: Equations Of Motion

We know that, under gravity the ball will trace out a parabola.

Set up the obvious coordinate axes with the ball struck at (0,0), reaching its peak height at (2.25,h), passing over the net at (3.6,1) and landing at (4.5,0).

A parabola is described by a quadratic equation, and it is not hard to see that in this case it must have the form

$y=kx(4.5-x) \dots(1)$

where k is some constant.

Substituting in the point where the ball crosses the net $(x,y)=(3.6,1)$ allows us to solve to find $k= \frac{1}{3.24}$

Substituting for k in equation (1) and setting $x=2.25$ allows us to find the maximum height of the ball

$h = \frac{1}{3.24} \times 2.25^{2}=1.5625$

Using the well known formula $t=\sqrt{\frac{2h}{g}}$ for the time taken for an object to drop from a height h, and remembering that in our problem the ball must also have risen to this height, we can find the answer for the total time of flight. Taking $g = 10 m/s^2$

$t= 2 \sqrt{\frac{2 \times 1.5625}{10}}=\boxed{1.118}$

Relevant wiki: Equations Of Motion

Let us denote the initial values of the horizontal and vertical component of the ball by $v_x$ and $v_y$ . The position of the ball as the function of time $t$ is

$x=v_x t$

$y=v_y t-\frac{g}{2} t^2$

where $x$ is the horizontal distance from the point when the ball was hit by the racket, and $y$ is the elevation about above court.

We have two points when we know $x$ and $y$ . First, when the ball clears the net it is at an elevation of $y_1=1$ m and the distance is $x_1=3.6$ m. Second, when the ball hits the court, $y_2=0$ m and $x_2=3.6+0.9=4.5$ m. Therefore we have from $x_1=v_x t_1$ and $y_1=v_y t_1-(g/2)t_1^2$ , $t_1=x_1/v_x$ and

$y_1=x_1 \frac{v_y}{v_x} -\frac{1}{2} g\left(\frac{x_1}{v_x}\right)^2$ .

On the other hand from $x_2=v_x t_2$ and $y_2=v_y t_2-(g/2)t_2^2$ , $t_2=x_2/v_x$ and

$y_2= 0 = x_2 \frac{v_y}{v_x} -\frac{1}{2} g\left(\frac{x_2}{v_x}\right)^2$

From this last equation we get $v_y=\frac{g x_2}{2 v_x}$ . We plug that in to the equation before that to get

$y_1=x_1 \frac{g x_2}{2 v_x^2} -\frac{1}{2} g\left(\frac{x_1}{v_x}\right)^2= \frac {g x_1(x_2-x_1)}{2 v_x^2}$ .

We re-arrange

$v_x= \sqrt{\frac{g x_1(x_2-x_1)}{2 y_1}}=4.02m/s$

The vertical component of the velocity is

$v_y=\frac{g x_2}{2 v_x}= 5.59m/s$ .

The time in the air is

$t_2=\frac{2 v_y}{g}= 1.12s$ .

Relevant wiki: 2D Kinematics Problem-solving

Let the ball be hit with an initial velocity $v$ at an angle of elevation $\theta$ with the horizontal. Then the horizontal displacement $x$ and vertical displacement $y$ in time $t$ are given by:

$\begin{cases} x = v \cos \theta \ t & ...(1) \\ y = v \sin \theta \ t - \dfrac 12 gt^2 &...(2) \end{cases}$

From $(1): \implies t = \dfrac x{v \cos \theta}$ and substituting $(2)$ , we have:

$\begin{aligned} y & = x \tan \theta - \frac {gx^2}{2v^2}\sec^2 \theta \quad \quad \small \color{#3D99F6} \text{when }x = 4.5, \ y = 0 \\ 0 & = \tan \theta - \frac {22.5}{v^2}\sec^2 \theta \quad \quad ...(3) \end{aligned}$

Since the trajectory is symmetrical, when $x=0.9$ , $y=1$ .

$\begin{aligned} 1 & = 0.9 \tan \theta - \frac {4.05}{v^2}\sec^2 \theta & ...(4) \end{aligned}$

$\begin{aligned} (4) \times \frac {50}9 - (3): \ \ \ \frac {50}9 & = 4 \tan \theta \implies \tan \theta = \frac {25}{18} \end{aligned}$

From $(3)$ :

$\begin{aligned} \tan \theta & = \frac {22.5}{v^2}\sec^2 \theta \\ \frac {25}{18} & = \frac {45}{2(v\cos \theta)^2} \\ \implies v\cos \theta & = \frac 9{\sqrt 5} \end{aligned}$

And time taken to travel $x = 4.5$ is

$\begin{aligned} v\cos \theta \ t & = 4.5 \\ \implies t & = 4.5 \times \frac {\sqrt 5}9 \approx \boxed{1.12} \text{ s} \end{aligned}$

This is a lot simpler than the solutions seem to suggest. Without air resistance we know that horizontal velocity is constant while with constant gravity, vertical velocity, v(t) is a linear function of time - in fact it is v-10t

1) The ball is in the air for T seconds so v = 5T

2) Since the trajectory is symmetric, the height after T/5 seconds is 1.0m.

The average vertical speed in this period is 4v/5 so 4v/5 × T/5 = 1.

Substituting for v, we get 4xT^2 =5 => T = sqrt (5)/2 = 1.12

When dealing with projectile motion, I like to analyze it from the perspective of the apex; i.e. choose $(0,0)$ at the top of the motion and $t = 0$ when the ball reaches this point. I will use downward as the positive $y$ -direction. The equations of motion are then $\begin{cases} x = \alpha t \\ y = \tfrac12 g t^2 = \beta x^2 & \beta = g/2\alpha^2. \end{cases}$

The endpoints have coordinate $E:\ (\pm r,h)$ ; the given point, $G:\ (s,h-y)$ . Here $r$ is half of the range, $s$ the horizontal distance of the given point from the center, and $y$ the given height above the ground. We have $\begin{cases} h = \beta r^2 \\ h - y = \beta s^2 \end{cases}$ $y = \beta (r^2 - s^2).$ Thus we are able to find the parameters, $\beta = \frac{y}{r^2-s^2},\ \ \ \alpha = \sqrt{\frac{g}{2\beta}} = \sqrt{\frac{g(r^2-s^2)}{2y}},$ and finally $t = \frac{2r}{\alpha} = 2r\sqrt{\frac{2y}{g(r^2 - s^2)}} = \sqrt{\frac{8y}{g(1-\gamma^2)}},$ where $\gamma = s/r$ .

Apply this with $y = 1$ , $g = 9.81$ , $\gamma = s/r = 1.35/2.25 = 3/5$ : $t = \sqrt{\frac{8\cdot 1}{9.81\cdot \left(1 - (\tfrac35)^2\right)}} = \sqrt{\frac{25}{2\cdot 9.81}} \approx \boxed{\SI{1.1288}{s}}.$

Decompose motion not into horizontal and vertical direction as usually the case, but into a constant speed linear motion along the tangent of the initial direction, and a simultaneous free fall h(t). At the moment T1=4t the ball reaches the net, and at T2=5t it hits the ground. The free fall are respectively 16h and 25h.

We have: 4s-16h=1; 5s-25h=0 for these two points, respectively. 4s and 5s are the upward distance traveled due to the constant speed linear motion.

Hence h=1/4, and 25h=25/4 is rhe tital free fall at the tine the ball reaches the ground. 25/4=gt^2/2. Thus t=1.12.

Since the path traced out by a projectile is given by the equation : $y = x \tan \theta$ ( $1- \frac{x}{R}$ ) (where $R$ is horizontal range and $x$ is horizontal displacement), we can use this equation at the point where the ball just crosses the net(i.e. when $y = 1$ and $x = 3.6$ ). Therefore -

$\Rightarrow$ $\frac {3.6}{5} \cdot \tan \theta = 1$ $\Rightarrow$ $\tan \theta = \frac{5}{3.6}$ = $\frac {1}{0.72}$

Also, the following relation between the time of flight $(T)$ and horizontal range $(R)$ holds for all projectiles :

$\frac{gT^2}{2} = R \tan \theta$ . We know that $R=4.5$ and $\tan \theta = \frac{1}{0.72}$ .We get -

$\Rightarrow$ $5T^2 = \frac {4.5}{0.72}$ $\Rightarrow$ $T^2$ = $\frac {0.9}{0.72} = \frac {5}{4}$ . Hence, the ball stays in the air for $T=\sqrt{\frac{5}{4}} \approx \boxed{1.118}$ seconds.

• The first and second equations used above can be derived by substituting $T=\frac{x}{v_0\cos \theta}$ into the equation for the vertical displacement(i.e.y= $v_0 \sin \theta \cdot T -\frac{gT^2}{2}$ ) and simplifying both sides of the equation using the formulas for $R$ and $T$ respectively. https://brilliant.org/wiki/projectile-motion-easy/?subtopic=kinematics&chapter=2d-kinematics

This is a projectile motion so we will be using the following formulas:

$V_{i} = \sqrt{\frac{R*g}{\sin(2\arctan (\frac{4.5}{3.24}))}}$

$V_{y} = V_{i} \sin \theta$

$T_{total} = \frac{2V_{y}}{g}$

First thing we need to know the value of is $\theta$

We know that the horizontal range is $R = 3.6 + 0.9 = 4.5$

The projectile can be represented as a graph. We know that it passes the x-axis at 0 and 4.5.

$f(0) = 0$ , $f(3.6) = 1$ and $f(4.5) = 0$ , as seen in the picture

and we get the equation $f(x) = a(x)(x - 4.5)$ then find $a$ to complete the equation. Insert $(3.6, 1)$ to the equation to find $a$

$1 = a(3.6)(3.6 -4.5)$

$a = \frac{1}{(3.6)(3.6 - 4.5)}$ or $\frac{1}{3.24}$

so plug $a$ back in

$f(x) = - \frac{1}{3.24} (x^2 - 4.5x)$

Where x is the horizontal distance and f(x) is the height.

then $f'(x) = - \frac{1}{3.24} (2x - 4.5)$ This is to find the slope of the tangent line at $(0, 0)$ and that will lead us to $\theta$

then the equation of the tangent line at $(0, 0)$ is $y = f'(0)x$ or $y = \frac{4.5}{3.24}x$

Then $\theta = \arctan (\frac{4.5}{3.24})$

Now that we have the value for $\theta$ , we can now find $V_{i}$

$V_{i} = \sqrt{\frac{R*g}{\sin(2\arctan (\frac{4.5}{3.24}))}}$

$V_{i} = \sqrt{\frac{4.5*10}{\sin(2\arctan (\frac{4.5}{3.24}))}}$

then get $V_{y}$ ,

$V_{y} = V_{i} \sin \theta$

$V_{y} = (\sqrt{\frac{4.5*10}{\sin(2\arctan (\frac{4.5}{3.24}))}}) (\sin (arctan (\frac{4.5}{3.24}))$

then get $T_{total}$

$T_{total} = \frac{2V_{y}}{g}$

$T_{total} = \frac{2(\sqrt{\frac{4.5*10}{\sin(2\arctan (\frac{4.5}{3.24}))}}) (\sin (arctan (\frac{4.5}{3.24}))}{10}$

$T_{total} = \frac{\sqrt{5}}{2}$ or $1.118033989...$ or $\boxed{1.12}$

The trajectory is a parabola that contains the following coordinates, where we consider the tennis court to be the x-axis, and the vertex of this porabola to be on the y-axis:

(-2.25,0) --> the ball's starting point, found by dividing the total horizontal by 2, which gives the distance from the origin $\frac{3.6+.9}{2}$ =2.25

(2.25,0) --> same as the previous point, except its the ball's end point on the other side of the net, which means its positive

(1.35,1) --> the tip of the net's height is 1, and the horizontal distance is the distance from the origin: 2.25 - .9 = 1.35

The general form of a quadratic equation: A $x^{2}$ + B x + C = y

By plugging in the points, we get a system of 3 equations of the coefficients:

A $(-2.25)^{2}$ + B (-2.25) + C = 0

A $(2.25)^{2}$ + B (2.25) + C = 0

A $(1.35)^{2}$ + B (1.35) + C = 0

From reduced row echelon form, we get that C = 1.5625, which is the vertex height. Now for elementary kinematics to calculate total time:

d = $v_{i}$ t + .5 a $t^{2}$

-1.5625 = 0(t) + .5 (10) $t^{2}$

t = .559017

For the total time, multiply by 2 to get the entire trajectory: $\boxed{t = 1.12}$

PS

Sorry, I'm new to Latex. I wish there were pre-done formatting options for the most commonly used functions, so that anyone can communicate their ideas much more quickly and much more efficiently with much less hassle of trying to learn the syntax of a new language. Its all about lessening the impact of technicalities of communication issues to prioritizing simply people sharing math ideas

this is just a helpbox.

1) in this problem use the formula for the relation between $y$ and $x$ in a projectile motion that is $y$ = $x$ $tan$ $\theta$ $-$ $gx^2$ $/$ $2v_0^2$ $cos^2$ $\theta$

2) Before using note that Range $(R)$ $=$ $4.5$ $m$ $=$ $v_0^2$ $sin$ $2$ $\theta$ $/$ $g$ = $2 v_0^2$ $sin$ $\theta$ $cos$ $\theta$ $/$ $g$ .

3) while using note that at $x = 3.6 m$ we have $y = 1 m$ .

4) Now, put values and at some stage you may have to manipulate the expression a bit by using point $(2)$ .

5) We are interested in $v_0$ $sin$ $\theta$ so that we can find the time of flight.Hence using point 3) we get the value of $v_0$ $cos$ $\theta$ .

6) Note that if we put this value into the Range formula we will get $v_0$ $sin$ $\theta$ and thus the answer. (1.118 seconds.)

We have to use the range formula twice once while solving for cos and once while solving for sin.