# Question cb891

Apr 19, 2017

This is a projectile motion question. The person leaves the ground with an initial velocity at an angle of ${28}^{\circ}$ to the horizontal.

This initial velocity has a $\textcolor{b l u e}{\text{horizontal vector component}}$ and a $\textcolor{b l u e}{\text{vertical vector component}}$.

You are asked to find the time of flight or total time, the total distance traveled and the peak height of the jumper. We will first start by figuring out the total time of the flight.

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To figure out the total time, we need to use one of the kinematic equations.

$\textcolor{w h i t e}{a a a a a a a a a a a a a a a}$$\textcolor{m a \ge n t a}{{v}_{f} = {v}_{o y} + a t}$

Where
${v}_{f y} = \text{final velocity at max height} \left(\frac{m}{s}\right)$
${v}_{o y} = \text{initial vertical velocity} \left(\frac{m}{s}\right)$
$a$ = acceleration due to gravity $\left(9.8 \frac{m}{s} ^ 2\right)$
$t = \text{time at max height (s)}$

Since acceleration due to gravity affects only the vertical velocity and we are solving for time using these variables, we will use only the vertical velocity component to solve for total time. Final velocity at max height is zero.

• ${v}_{f y} = {v}_{o y} + a t$
• $0 = \left(12 \sin {28}^{\circ}\right) + \left(- 9.8\right) t$
• $\frac{0 - 5.63}{-} 9.8 = t$
• $0.57 s = t$

Since we figured out the time at max height, this is taken to be half the flight. Since it takes the same amount of time to reach max height as it does to fall back down, we just multiply the time we just figured out by 2 to get the total time for the whole flight.

$0.57 s \cdot 2 = \textcolor{b l u e}{1.14 s}$

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To find the total distance traveled by the person, we will use the following equation:

$\textcolor{w h i t e}{a a a a a a a a a a a a a}$$\textcolor{m a \ge n t a}{\text{velocity} = \frac{\Delta \mathrm{di} s p l a c e m e n t}{\Delta t i m e}}$

Since we are finding displacement in the x direction, and we need velocity in order to solve it, we will use the horizontal velocity component. Time is taken to equal to the total time of the flight, $1.14 s$

• ${v}_{\text{ox") = (Delta "displacement")/(Delta "time}}$

• v_("ox") * Delta "time"= Delta "displacement"#

• $\left(12 \cos {28}^{\circ}\right) \cdot 1.14 \text{ = Delta "displacement}$

• $10.59 \cdot 1.14 = \Delta \text{displacement}$

• $\textcolor{b l u e}{12.07 m = \Delta \text{displacement}}$

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To find the peak height, or max height, we will use the following equation:

$\textcolor{w h i t e}{a a a a a a a a a a a a a a a}$$\textcolor{m a \ge n t a}{t = \sqrt{\frac{2 h}{g}}}$

Where
$t = \text{time at max height}$
$h = \text{max height in y direction}$
$g = \text{acceleration due to gravity} \left(9.8 \frac{m}{s} ^ 2\right)$

Plugin, isolate h and solve.

• $t = \sqrt{\frac{2 h}{g}}$

• $\left(0.57\right) = \sqrt{\frac{2 h}{9.8}}$

• ${\left(0.57\right)}^{2} = {\left(\sqrt{\frac{2 h}{9.8}}\right)}^{2}$

• ${\left(0.57\right)}^{2} = \frac{2 h}{9.8}$

• $\frac{\left(9.8\right) {\left(0.57\right)}^{2}}{2} = h$

• $\textcolor{b l u e}{h = 1.59 m}$