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The scenario: A diver leaves a $3m$ board on a trajectory that takes her $2.3 m$ above the board and then into the water 2.9 m horizontally from the end of the board.

The questions:

a) At what speed did she leave the board?

b) At what angle did she leave the board?

My attempt at this problem:

First I found an expression for time when velocity is $0$ (at maximum $y$ displacement of trajectory) using $V_y = V_{y_0} + a\cdot T$, therefore

$$0 = V\cdot sin(\theta)-9.8\cdot T$$ and so $$T = v\cdot \frac{sin(\theta)}{9.8}$$

Then I found an expression for initial velocity when vertical displacement is $2.3$ meters using $S_y = V_y + \frac{1}{2}\cdot a\cdot T$, therefore

$$2.3 = V\cdot \sin(\theta)\cdot V\cdot \frac{\sin(\theta)}{9.8} - 4.9\cdot \left(V\cdot \frac{\sin(\theta)}{9.8}\right)^2$$

$$2.3 = V^2\cdot \frac{\sin^2(\theta)}{9.8} - 4.9\cdot V^2 \frac{\sin^2(\theta)}{96.04}$$

$$2.3 + 4.9\cdot V^2 \cdot \frac{sin^2(\theta)}{96.04} = V^2\cdot \frac{sin^2(\theta)}{9.8}$$

$$220.892 + 4.9\cdot V^2 \cdot \sin^2(\theta) = 9.8\cdot V^2\cdot \sin^2(\theta)$$

$$220.892 = 4.9\cdot V^2 \cdot \sin^2(\theta)$$

$$V^2 = \frac{220.892}{4.9\cdot sin^2(\theta)}$$

$$V = \sqrt{\frac{45.08}{sin^2(\theta)}}$$

Next I attempted to calculate the angle at which the diver left the board using $S_x = V_{x_0}\cdot T$, with the horizontal displacement being half of total horizontal displacement $\left(\frac{2.9}{2} = 1.45m\right)$ at the time when maximum vertical displacement is $2.3$ meters, therefore

$$1.45 = V\cdot cos(\theta)\cdot T \qquad \text{(using T and V from previous two steps)}$$

$$1.45 = \sqrt{\frac{45.08}{\sin^2(\theta)}}\cdot \cos(\theta) \cdot\sqrt{\frac{45.08}{\sin^2(\theta)}}\cdot \frac{\sin(\theta)}{9.8}$$

$$2.1025 = \frac{45.08}{\sin^2(\theta)}\cdot \cos^2(\theta) \cdot \frac{45.08}{\sin^2(\theta)} \cdot \frac{\sin^2(\theta)}{96.04}$$

$$2.1025 = 45.08\cdot \frac{\cos^2(\theta)}{\sin^2(\theta)} \cdot \frac{45.08 \cdot \frac{\sin^2(\theta)}{\sin^2(\theta)}}{96.04} \qquad [\sin^2(\theta)/\sin^2(\theta) = 1]$$

$$2.1025 = 45.08\cdot \frac{1}{\tan(\theta)} \cdot \frac{45.08}{96.04}$$

$$2.1025 = \frac{21.16}{\tan(\theta)}$$

$$\theta = tan^{-\frac{21.16}{2.1025}}$$

$$\theta = 84.33 \text{ degrees}$$

Then I substituted the calculated angle back into the expression I found for initial velocity.

$$V = \sqrt{\frac{45.08}{sin^2(\theta)}}$$ $$V = \sqrt{\frac{45.08}{sin^2(84.33)}}$$ $$V = 6.747 \frac{m}{s}$$

As I submitted these answers, they were both incorrect and I am not sure why.

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  • $\begingroup$ The equations look quite messy, I would recommend using MathJax. $\endgroup$ – Matti P. Mar 27 '18 at 7:55
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Let $t_1$ and $t_2$ be the times elapsed before and after the highest point of trajectory, and $h$, $h_1$ and $d$ be the height of the board, that of the highest point and the horizontal displacement, respectively. Then: $$\begin{align} &v_x(t_1+t_2)=d,\\ &v_y=at_1,\\ &\frac{at_1^2}{2}=h_1,\\ &\frac{at_2^2}{2}=h+h_1. \end{align} $$ Solving the system of equations one obtains: $$ \begin{align} &v_x=\frac{d\sqrt{a}}{\sqrt{2(h+h_1)}+\sqrt{2h_1}},\\ &v_y=\sqrt{2ah_1}. \end{align} $$

Can you take it from here?

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  • $\begingroup$ I'm able to solve the system of equations that you have provided to find the value of Vx as 1.681 m/s and Vy as 6.714 m/s which gives an initial velocity of 6.92 m/s. This gives a correct answers, however i'm confused on how you got those equations. $\endgroup$ – Dani Green Mar 27 '18 at 9:44
  • $\begingroup$ I originally misunderstood the part of the question which mentioned a 3m board. For some reason i assumed that it was referring to the board being 3m long rather than starting at a height of 3m. I'll attempt the problem again with this in mind. $\endgroup$ – Dani Green Mar 27 '18 at 9:46
  • $\begingroup$ @Dani Green With this in mind do you have still questions about the derivation of the result? $\endgroup$ – user Mar 27 '18 at 9:52
  • $\begingroup$ Yes, I still dont understand why you used the equations that you did. Have you only written parts of the equations that you used, because some of the bits look similar to the kinematic equations? $\endgroup$ – Dani Green Mar 27 '18 at 10:03
  • $\begingroup$ I used the fact that $V_y=0$ at the highest point. Do you know that in this case $V=at$ and $S=at^2/2$? Otherwise please tell more precisely what is still unclear. $\endgroup$ – user Mar 27 '18 at 10:11

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