# A projectile is fired horizontally from a gun that is 45.0 m above fla

A projectile is fired horizontally from a gun that is 45.0 m above flat ground, emerging from the gun with a speed of 250 $\frac{m}{s}$.
a) How long does the projectile remain in the air?
b) At what horizontal distance from the firing point does it strike the ground?
c) What is the magnitude of the vertical component of its velocity as it strikes the ground?
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Wendy Boykin

Givens:
${y}_{0}=45m$
${v}_{x}=250\frac{m}{s}$
Part a:
To find the time it stays in in the air we use this equation
$y={y}_{0}+{v}_{0y}t-\frac{1}{2}{gt}^{2}$
Where we set our coordinates so that the bullet starts at vertical position $y=0$ and velocity ${v}_{0y}=0$.
Therefore
$0=\left(45m\right)-\frac{1}{2}\left(9.8\frac{m}{s}\right){t}^{2}$
$t=\sqrt{\frac{2\left(45m\right)}{\left(9.8\frac{m}{s}\right)}}=3.03s$
Part b:
To find the horizontal distance from the firing point at which it strikes the ground we use the equation
$x={x}_{0}+{v}_{x}t$
Where we use the time t we obtained previously and we choose our coordinates so that the particle starts at position ${x}_{0}=0$
$x={x}_{0}+{v}_{x}t=0+\left(250\frac{m}{s}\right)\left(3.03s\right)=757.5m$
Part c:
Here we need to know the magnitude of ${v}_{y}$ at time $t=3.03s$ We make use of:
${v}_{y}={v}_{0y}-gt$
To find the velocity ${v}_{y}$ at time of impact with the ground
${v}_{y}=-\left(9.8\frac{m}{s}\right)\left(3.03s\right)=29.694\frac{m}{s}$

mauricio0815sh

I make gravitational acceleration to be positive, because as it hits the ground, the bullet is gaining speed; not losing.
$0={\left[\left(250\frac{m}{s}\right)\left(\mathrm{sin}\theta \right)\right]}^{2}+2\left(9.8\frac{m}{{s}^{2}}\right)\left(0m-45m\right)$
$882\frac{{m}^{2}}{{s}^{2}}=\left(62500\frac{{m}^{2}}{{s}^{2}}\right)\left(\mathrm{sin}\theta \right)$
$\mathrm{sin}\theta =-0.1188$
I make the value of $\mathrm{sin}\theta$ to be negative, because I'm making the gun to face the rightward direction. If it shoots, then it goes down, from the path driven by its velocity.
$\theta =-6.823°$
${v}_{y}={v}_{0}\mathrm{sin}\theta -gt$
$0=\left(250\frac{m}{s}\right)\left(\mathrm{sin}\left(-6.823°\right)\right)+\left(9.8\frac{m}{{s}^{2}}\right)\left(t\right)-\left(9.83\frac{m}{{s}^{2}}\right)\left(t\right)=\left(-29.7\frac{m}{s}\right)$
$t=3.03s$
Then here's where I'm having conflicting numbers, when | calculate the horizontal distance.
$x-{x}_{0}=\left({v}_{0}\mathrm{cos}\theta \right)t$
$x=\left(250\frac{m}{s}\right)\left(\mathrm{cos}\left(-6.823°\right)\right)\left(3.03s\right)=752m\ne 758m$