 ROFHIWA NDOU

2022-07-03

A projectile is fired with an initial speed of 36.6 m/s at an angle of 42.2 above the horizontal on a long flat firing range. Determine (a) the maximum height reached by the projectile, (b) the total time in the air, (c) the total horizontal distance covered (that is, the range), and (d) the speed of the projectile 1.50 s after firing user_27qwe

(a) To determine the maximum height reached by the projectile, we can use the fact that at the highest point of its trajectory, the vertical velocity component becomes zero. We can find the time it takes to reach the highest point using the vertical motion equation:
$y={v}_{0}\mathrm{sin}\left(\theta \right)t-\frac{1}{2}g{t}^{2}$
where:
$y$ is the vertical displacement,
${v}_{0}$ is the initial speed,
$\theta$ is the launch angle,
$t$ is the time,
and $g$ is the acceleration due to gravity.
At the highest point, the vertical displacement $y$ is the maximum height reached, and the vertical velocity component becomes zero. Therefore, we can set ${v}_{y}=0$ and solve for $t$:
$0={v}_{0}\mathrm{sin}\left(\theta \right)-gt$
Solving for $t$:
$t=\frac{{v}_{0}\mathrm{sin}\left(\theta \right)}{g}$
Now, we can substitute the given values into the equation:
${v}_{0}=36.6\phantom{\rule{0.167em}{0ex}}\text{m/s}$
$\theta ={42.2}^{\circ }$
$g=9.8\phantom{\rule{0.167em}{0ex}}{\text{m/s}}^{2}$
Substituting these values into the equation, we can calculate $t$:
$t=\frac{36.6\phantom{\rule{0.167em}{0ex}}\text{m/s}·\mathrm{sin}\left({42.2}^{\circ }\right)}{9.8\phantom{\rule{0.167em}{0ex}}{\text{m/s}}^{2}}$
Calculating this expression, we find:
$t\approx 2.68\phantom{\rule{0.167em}{0ex}}\text{s}$
Now, we can find the maximum height $H$ by substituting the calculated time into the vertical motion equation:
$y={v}_{0}\mathrm{sin}\left(\theta \right)t-\frac{1}{2}g{t}^{2}$
$H=36.6\phantom{\rule{0.167em}{0ex}}\text{m/s}·\mathrm{sin}\left({42.2}^{\circ }\right)·2.68\phantom{\rule{0.167em}{0ex}}\text{s}-\frac{1}{2}·9.8\phantom{\rule{0.167em}{0ex}}{\text{m/s}}^{2}·\left(2.68\phantom{\rule{0.167em}{0ex}}\text{s}{\right)}^{2}$
Calculating this expression, we find:
$H\approx 46.8\phantom{\rule{0.167em}{0ex}}\text{m}$
Therefore, the maximum height reached by the projectile is approximately 46.8 meters.
(b) To determine the total time in the air, we need to consider the entire trajectory of the projectile, which includes both the ascent and descent. The total time in the air, $T$, can be found by multiplying the time taken to reach the maximum height by 2:
$T=2t$
Substituting the value of $t$ we calculated earlier:
$T=2·2.68\phantom{\rule{0.167em}{0ex}}\text{s}$
Calculating this expression, we find:
$T\approx 5.36\phantom{\rule{0.167em}{0ex}}\text{s}$
Therefore, the total time in the air is approximately 5.36 seconds.
(c) To determine the total horizontal distance covered (range), we can use the horizontal motion equation:
$x={v}_{0}\mathrm{cos}\left(\theta \right)t$
where $x$ is the horizontal distance traveled.
Substituting the given values into the equation:
${v}_{0}=36.6\phantom{\rule{0.167em}{0ex}}\text{m/s}$
$\theta ={42.2}^{\circ }$
$t=5.36\phantom{\rule{0.167em}{0ex}}\text{s}$
$x=36.6\phantom{\rule{0.167em}{0ex}}\text{m/s}·\mathrm{cos}\left({42.2}^{\circ }\right)·5.36\phantom{\rule{0.167em}{0ex}}\text{s}$
Calculating this expression, we find:
$x\approx 172.2\phantom{\rule{0.167em}{0ex}}\text{m}$
Therefore, the total horizontal distance covered (range) is approximately 172.2 meters.
(d) To determine the speed of the projectile 1.50 seconds after firing, we can use the horizontal and vertical motion equations. At any given time, the speed of the projectile can be calculated using the magnitude of its velocity vector:
$v=\sqrt{{{v}_{x}}^{2}+{{v}_{y}}^{2}}$
where ${v}_{x}$ is the horizontal component of velocity and ${v}_{y}$ is the vertical component of velocity.
The horizontal component of velocity remains constant throughout the motion. Therefore, ${v}_{x}={v}_{0}\mathrm{cos}\left(\theta \right)$.
The vertical component of velocity can be found using the equation:
${v}_{y}={v}_{0}\mathrm{sin}\left(\theta \right)-gt$
Substituting the given values into the equations:
${v}_{0}=36.6\phantom{\rule{0.167em}{0ex}}\text{m/s}$
$\theta ={42.2}^{\circ }$
$g=9.8\phantom{\rule{0.167em}{0ex}}{\text{m/s}}^{2}$
$t=1.50\phantom{\rule{0.167em}{0ex}}\text{s}$
We can calculate the horizontal and vertical components of velocity:
${v}_{x}=36.6\phantom{\rule{0.167em}{0ex}}\text{m/s}·\mathrm{cos}\left({42.2}^{\circ }\right)$
${v}_{y}=36.6\phantom{\rule{0.167em}{0ex}}\text{m/s}·\mathrm{sin}\left({42.2}^{\circ }\right)-9.8\phantom{\rule{0.167em}{0ex}}{\text{m/s}}^{2}·1.50\phantom{\rule{0.167em}{0ex}}\text{s}$
Calculating these expressions, we find:
${v}_{x}\approx 26.61\phantom{\rule{0.167em}{0ex}}\text{m/s}$
${v}_{y}\approx 12.08\phantom{\rule{0.167em}{0ex}}\text{m/s}$
Finally, we can calculate the speed of the projectile 1.50 seconds after firing:
$v=\sqrt{{{v}_{x}}^{2}+{{v}_{y}}^{2}}$
$v=\sqrt{\left(26.61\phantom{\rule{0.167em}{0ex}}\text{m/s}{\right)}^{2}+\left(12.08\phantom{\rule{0.167em}{0ex}}\text{m/s}{\right)}^{2}}$
Calculating this expression, we find:
$v\approx 29.18\phantom{\rule{0.167em}{0ex}}\text{m/s}$
Therefore, the speed of the projectile 1.50 seconds after firing is approximately 29.18 m/s.

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