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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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?

Wendy Boykin · Accepted Answer

Givens: y0=45m vx=250ms  Part a: To find the time it stays in in the air we use this equation y=y0+v0yt−12gt2 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=v0y−gt To find the velocity vy at time of impact with the ground vy=−(9.8ms)(3.03s)=29.694ms

mauricio0815sh · Accepted Answer

I make gravitational acceleration to be positive, because as it hits the ground, the bullet is gaining speed; not losing. 0=[(250ms)(sin⁡θ)]2+2(9.8ms2)(0m−45m) 882m2s2=(62500m2s2)(sin⁡θ) sin⁡θ=−0.1188 I make the value of sin⁡θ to be negative, because I&#039;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=v0sin⁡θ−gt 0=(250ms)(sin⁡(−6.823°))+(9.8ms2)(t)−(9.83ms2)(t)=(−29.7ms) t=3.03s Then here&#039;s where I&#039;m having conflicting numbers, when | calculate the horizontal distance. x−x0=(v0cos⁡θ)t x=(250ms)(cos⁡(−6.823°))(3.03s)=752m≠758m

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

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