# Calculating initial velocity and angle given displacement

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.

• The equations look quite messy, I would recommend using MathJax. – Matti P. Mar 27 '18 at 7:55

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}
• 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. – user Mar 27 '18 at 10:11