If a projectile is shot at a velocity of $2 \frac{m}{s}$ $2 m/s$ and an angle of $\frac{π}{4}$ $pi/4$ , how far will the projectile travel before landing?

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If a projectile is shot at a velocity of $2 \frac{m}{s}$ $2 m/s$ and an angle of $\frac{π}{4}$ $pi/4$ , how far will the projectile travel before landing?

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Answer 1 · 2016-01-05T18:57:49

it will land about 0.638 m far.

Explanation:

We have to write down the equations of motion:

$x = x_{0} + v cos θ t$ $x= x_0+vcosthetat$
$y = h_{0} + v sin θ t + \frac{a t^{2}}{2}$ $y= h_0+ v sintheta t + (at^2)/2$

where $v$ $v$ is the velocity of the projectile, $θ$ $theta$ is the angle of the trajectory, $h_{0}$ $h_0$ the initial quota of the projectile, $x_{0}$ $x_0$ its initial position.

We have:

$h_{0} = 0$ $h_0=0$ and $x_{0} = 0$ $x_0=0$ ;
$v=2 \ ms^(-1)$
$theta= pi/4$
$a=-g \ ~~ -9.81 \ ms^(-2).$

We have $v$ and $thetha$ : we have to find t from the first equation and substitute its value in the second one.
When the projectile lands, it reaches the ground so we have y=0 at that moment.
The equations of motion at the landing are:

$x= vcosthetat$
$0= v sintheta t - (gt^2)/2$

from the first equation $t=x/(vcostheta)$ .
The other becomes:
$v sin theta * x/(vcostheta) - g/2*(x/(vcostheta))^2=$

Performing the calculation we eventually have:
$x= sqrt( (2v^2 costheta sintheta)/g )~~ 0.638 \ m$

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If a projectile is shot at a velocity of $2 \frac{m}{s}$ $2 m/s$ and an angle of $\frac{π}{4}$ $pi/4$ , how far will the projectile travel before landing?

1 Answer

Explanation:

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