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

David Lewis 2021-12-16 Answered
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 ms.
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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Expert Answer

Wendy Boykin
Answered 2021-12-17 Author has 35 answers

Givens:
y0=45m
vx=250ms
Part a:
To find the time it stays in in the air we use this equation
y=y0+v0yt12gt2
Where we set our coordinates so that the bullet starts at vertical position y=0 and velocity v0y=0.
Therefore
0=(45m)12(9.8ms)t2
t=2(45m)(9.8ms)=3.03s
Part b:
To find the horizontal distance from the firing point at which it strikes the ground we use the equation
x=x0+vxt
Where we use the time t we obtained previously and we choose our coordinates so that the particle starts at position x0=0
x=x0+vxt=0+(250ms)(3.03s)=757.5m
Part c:
Here we need to know the magnitude of vy at time t=3.03s We make use of:
vy=v0ygt
To find the velocity vy at time of impact with the ground
vy=(9.8ms)(3.03s)=29.694ms

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mauricio0815sh
Answered 2021-12-18 Author has 34 answers

I make gravitational acceleration to be positive, because as it hits the ground, the bullet is gaining speed; not losing.
0=[(250ms)(sinθ)]2+2(9.8ms2)(0m45m)
882m2s2=(62500m2s2)(sinθ)
sinθ=0.1188
I make the value of sinθ 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.
θ=6.823°
vy=v0sinθgt
0=(250ms)(sin(6.823°))+(9.8ms2)(t)(9.83ms2)(t)=(29.7ms)
t=3.03s
Then here's where I'm having conflicting numbers, when | calculate the horizontal distance.
xx0=(v0cosθ)t
x=(250ms)(cos(6.823°))(3.03s)=752m758m

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