End of chapter exercises

is always downwards

A. The same velocity and the same acceleration.

b) \({u}_{x}={u}_{y}\)

d) displacement

c) the same as when it left the throwers hand.

C - the force is constant since the acceleration is constant.

The ball will travel upwards, stop at its highest point, and then fall back down to the ground. As it reaches the initial position of \(\text{5}\) metres above the ground, it will be travelling once again at \(\text{10}\) \(\text{m·s$^{-1}$}\) but this time in the downwards direction. Taking downwards as the +ve direction:

\begin{align*} v_{f}^2 & = v_{i}^2 + 2a\Delta x \\ &= (\text{10})^2 + 2(\text{9,8})(\text{5}) \\ &= \text{198} \\ v_{f} &= \sqrt{\text{198}} = \text{14,07}\text{ m·s$^{-1}$} \end{align*}

1. \(\text{5}\) \(\text{m·s$^{-1}$}\)

2. the acceleration of the ball

3. \begin{align*} \text{gradient} &= \frac{\Delta y}{\Delta x} \\ &= \frac{(\text{5}- (-\text{15}))}{(\text{2}- \text{0})} \\ & = \text{10} \end{align*} \(\text{10}\) \(\text{m·s$^{-2}$}\)

4. According to the graph, the ball reaches the boy's hand again after \(\text{1}\) second (when the downward speed of \(\text{5}\) \(\text{m·s$^{-1}$}\) is the same as its initial upward speed.) Therefore the area under the graph between \(\text{1}\) and \(\text{2}\) seconds (when it reaches its maximum negative velocity) is equal to the remaining distance the ball falls to the ground. The area is equal to \(\text{10}\).

   Distance = \(\text{10}\) \(\text{m}\).

5. It takes \(\text{2}\) \(\text{s}\) until the ball bounces. Calculate the additional time from bouncing to reach maximum height. Assume the acceleration is the same as before, i.e. \(\text{10}\) \(\text{m·s$^{-2}$}\).

   \begin{align*} v_{f} & = v_{i} + at \\ t & = \frac{v_{f} - v_{i}}{a} \\ & = \frac{\text{0}-\text{10}}{-\text{10}} \\ & = \text{1}\text{ s} \end{align*}

   Therefore the total time taken is \(\text{3}\) \(\text{s}\).

6. Calculate the area under the graph for the final \(\text{1}\) second or use equation of motion to find maximum height after the bounce:

   \begin{align*} \Delta x &= v_{i}t + \frac{1}{2}at^2 \\ &= (\text{10})(\text{1}) + \frac{1}{2}(-\text{10})(\text{1})^2 \\ & = \text{5}\text{ m} \end{align*}

   Therefore the position of the ball, measured from the boy's hand is\(\text{5}\) \(\text{m}\)-\(\text{5}\) \(\text{m}\) = \(\text{0}\) \(\text{m}\). In other words, it is the same height as the boy's hand.

1. Calculate the area under the graph to find the total height she falls. For the first \(\text{4}\) \(\text{s}\), approximate area with a triangle:

   \begin{align*} \Delta x_{1} & = \frac{1}{2}\text{base}\times\text{height} \\ &= \frac{1}{2}(\text{4})(\text{40}) \\ &= \text{80}\text{ m} \end{align*}

   For the time \(\text{4}\)-\(\text{8}\) \(\text{s}\):

   \begin{align*} \Delta x_{2} & = \text{base}\times\text{height} \\ & = (\text{4})(\text{40}) \\ & = \text{160}\text{ m} \end{align*}

   For the time \(\text{8}\)-\(\text{9}\) \(\text{s}\), approximate area with a triangle plus a square:

   \begin{align*} \Delta x_{3} & = \frac{1}{2}\text{base}\times\text{height} + \text{base}\times\text{height}_2\\ & = \frac{1}{2}(1)(\text{35}) + (\text{1})(\text{5})\\ & = \text{22,5}\text{ m} \end{align*}

   For the time \(\text{9}\)-\(\text{15}\) \(\text{s}\), approximate area with a rectangle:

   \begin{align*} \Delta x_{4} & = (6)(5) \\ & =\text{30}\text{ m} \end{align*}

   For the time \(\text{15}\)-\(\text{16}\) \(\text{s}\), approximate area with a triangle:

   \begin{align*} \Delta x_{5} & = \frac{1}{2}(1)(5) \\ & = \text{2,5}\text{ m} \end{align*}

   Total distance fallen = \(\text{80}\text{ m} + \text{160}\text{ m} + \text{22,5}\text{ m} + \text{30}\text{ m} + \text{2,5}\text{ m} = \text{295}\text{ m}\)

   .

   Rounded to the nearest \(\text{10}\) \(\text{m}\) is \(\text{300}\) \(\text{m}\)

2. \(\text{5}\) \(\text{m·s$^{-1}$}\)

3. \(\text{600}\) \(\text{N}\)

4. Once the parachute is open, the parachutists falls at constant speed, implying that her net acceleration is zero. Therefore the force on her due to gravity and the force on her due to air resistance are equal and opposite:

   \begin{align*} \vec{F}_g & = \vec{F}_{air} \\ mg &= \text{600}\text{ N} \\ m(\text{9,8}) &= \text{600}\text{ N} \\ m & = \text{61,22}\text{ kg} \end{align*}

1. \(\text{300}\) \(\text{m}\)

2. \(\text{5}\) \(\text{m·s$^{-1}$}\)

3. \(\text{600}\) \(\text{N}\)

4. \(\text{61,22}\) \(\text{kg}\)

1. \begin{align*} v_{f}^2 & = v_{i}^2 + 2a\Delta x \\ \Delta x & = \frac{v_{f}^2 - v_{i}^2}{2a} \\ & = \frac{(\text{0})^2-(\text{13})^2}{\text{2}(-\text{9,8})} \\ & = \text{8,62}\text{ m} \end{align*}

2. Calculate the time taken for the ball to reach its maximum height:

   \begin{align*} v_{f} & = v_{i} + at \\ t & = \frac{ v_{f}-v_{i}}{a} \\ & = \frac{\text{0}-\text{13}}{-\text{9,8}} \\ & = \text{1,33}\text{ s} \end{align*}

   Now calculate the acceleration needed:

   \begin{align*} \Delta x & = v_{i}t + \frac{1}{2}at^{2} \\ & = (\text{0})t + \frac{1}{2}at^{2} \\ a & = \frac{2\Delta x}{t^2} \\ & = \frac{2(\text{10} - \text{8,62})}{(\text{1,33})^2} = \text{1,63}\text{ m·s$^{-2}$} \end{align*}

1. \(\text{8,62}\) \(\text{m}\)

2. \(\text{1,63}\) \(\text{m·s$^{-2}$}\)