# Question c6d6a

Jun 14, 2015

Time of flight $t = \text{2.44 s}$
Maximum vertical height $h = \text{7.32 m}$
Horizontal range ${d}_{x} = \text{45.0 m}$

#### Explanation:

Your projectile is fired with an initial velocity of 22 m/s and at an angle of ${33}^{\circ}$ with the horizontal.

${v}_{0} = \text{22 m/s}$

$\theta = {33}^{\circ}$ This means that the motion of the projectile has a vertical component and a horizontal component.

Vertically, the projectile will be acted upon by the gravitational acceleration, $g$. This means that you can write

${d}_{y} = {v}_{0 y} \cdot t - \frac{1}{2} \cdot g \cdot {t}^{2}$, where

${d}_{y}$ - the total vertical displacement;
${v}_{0 y}$ - the vertical component of ${v}_{0}$;
$t$ - the total time of flight.

In your case ${d}_{y}$ will be equal to zero because the projectile starts from the ground level and ends up on the ground level. The vertical component of ${v}_{0}$ will be equal to

${v}_{0 y} = {v}_{0} \cdot \sin \theta$

The equation becomes

$0 = {v}_{0} \cdot \left(\sin \theta\right) \cdot t - \frac{1}{2} \cdot g \cdot {t}^{2}$

$- \frac{1}{2} \cdot 9.8 {t}^{2} + 22 \sin \left({33}^{\circ}\right) t = 0$

$- 4.9 {t}^{2} + 11.98 t = 0 \iff t \left(- 4.9 t + 11.98\right) = 0$

The only acceptable solution is

$t = \frac{- 11.98}{-} 4.9 = \textcolor{g r e e n}{\text{2.44 s}}$

At the maximum vertical height, the vertical component of the velocity is zero. This means that you can write

$0 = {v}_{0 y}^{2} - 2 \cdot g \cdot h$, where

$h$ - the maximum height the projectile reaches.

Solving this equation for $h$ will get you

$h = \frac{{v}_{0}^{2} \sin {\left({33}^{\circ}\right)}^{2}}{g} = \textcolor{g r e e n}{\text{7.32 m}}$

Horizontally, the projectile is not acted upon by any forces. This means that you can write

${d}_{x} = {v}_{0 x} \cdot t$, where

${d}_{x}$ - the range of the projectile;
${v}_{\otimes}$ - the horizontal component of ${v}_{0}$.

Use the values you have to solve for ${d}_{x}$

d_x = v_0 * costheta * t = 22"m"/cancel("s") * cos(33^@) * 2.44cancel("s")#

${d}_{x} = \textcolor{g r e e n}{\text{45.0 m}}$

Jun 14, 2015

1) time of flight =2.44sec 2) maximum vertical height= 7.30 metre
3)horizontal range= 82.6

#### Explanation:

it is defined as the time taken by any object to travel in any medium it is calculated by

t=2vsintheta/a
maximum vertical height is the peak of that projectile motion the object is travelling
horizontal range it is the range of horizontal motion which is done by object