Question

A diver springs upward from a board that is 4.50 m above the water. At the instant she contacts the water her speed is 15.4 m/s and her body makes an angle of 66.0 ° with respect to the horizontal surface of the water. Determine her initial velocity, both (a) magnitude and (b) direction.

Answer 1

A diver springs upward from a board that is $4.50\,\mathrm{m}$ above the water. At the instant she contacts the water her speed is $15.4\,\mathrm{m/s}$ and her body makes an angle of $66.0{}^{\circ}$ with respect to the horizontal surface of the water. Determine her initial velocity, both (a) magnitude and (b) direction.

**Solution:**

$\beta$ – given angle $66{}^{\circ}$

The Y (vertical) component of the velocity is

$$\sin 66{}^{\circ} \cdot 15.4\,\frac{\mathrm{m}}{\mathrm{s}} = 14.07\,\frac{\mathrm{m}}{\mathrm{s}}$$

The X (horizontal) component of the velocity is

$$\cos 66.0{}^{\circ} \cdot 15.4\,\frac{\mathrm{m}}{\mathrm{s}} = 6.26\,\frac{\mathrm{m}}{\mathrm{s}}$$

The horizontal component of the velocity is the same on take-off.

The angle is the arc tangent of the vertical component over the horizontal component.

Rate equation for the diver along Y-axis:

$$V(y) = gt, \, g = 9.8\,\frac{\mathrm{m}}{\mathrm{s}^2}, \, t - \text{time from the top of the flight}$$ $$t = \frac{14.07\,\frac{\mathrm{m}}{\mathrm{s}}}{9.8\,\frac{\mathrm{m}}{\mathrm{s}^2}} = 1.44\,\mathrm{sec}$$

$h$ – height at the top of the flight to the water:

$$\mathrm{y: } \mathrm{h} = \frac{\mathrm{g t}^{2}}{2} = \frac{9.8 \frac{\mathrm{m}}{\mathrm{s}^{2}} \cdot (1.44 \mathrm{s})^{2}}{2} = 10.2 \mathrm{m}$$

d - height at the top of the flight to the board:

$$\mathrm{d} = 10.2 \mathrm{m} - 4.5 \mathrm{m} = 5.7 \mathrm{m}$$

$\mathrm{d} = \frac{1}{2} \mathrm{g T}^{2}$, where $\mathrm{T}$ - the time to the top of the flight from the board.

$$\mathrm{T} = \sqrt{\frac{2 \mathrm{d}}{\mathrm{g}}} = \sqrt{\frac{2 \cdot 5.7 \mathrm{m}}{9.8 \frac{\mathrm{m}}{\mathrm{s}^{2}}}} = 1.08 \mathrm{sec}$$ $$V_{y\text{-init}} = \mathrm{g T} = 9.8 \frac{\mathrm{m}}{\mathrm{s}^{2}} \cdot 1.08 \mathrm{sec} = 10.6 \frac{\mathrm{m}}{\mathrm{s}}$$ $$V_{x\text{-init}} = \cos 66.0{}^{\circ} \cdot 15.4 \frac{\mathrm{m}}{\mathrm{s}} = 6.26 \frac{\mathrm{m}}{\mathrm{s}}$$ $$V_{0} = \sqrt{V_{x\text{-init}}^{2} + V_{y\text{-init}}^{2}} = \sqrt{\left(10.6 \frac{\mathrm{m}}{\mathrm{s}}\right)^{2} + \left(6.26 \frac{\mathrm{m}}{\mathrm{s}}\right)^{2}} = 12.31 \frac{\mathrm{m}}{\mathrm{s}}$$ $$\text{Angle} = \arctan \frac{V_{y\text{-init}}}{V_{x\text{-init}}} = \arctan \frac{10.6 \frac{\mathrm{m}}{\mathrm{s}}}{6.26 \frac{\mathrm{m}}{\mathrm{s}}} = 59.44 \text{ degrees}$$

Answer: (a) $V_{0} = 12.31 \frac{\mathrm{m}}{\mathrm{s}}$

(b) Angle $= \alpha = 59.44$ degrees