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A projectile is shot from the edge of a cliff h = 245 m above ground level with an initial speed of v0 = 145 m/s at an angle of 37.0° with the horizontal

how long would it take the projectile to hit point P at ground level.
s
How would I find the range X of the projectile motion, would it be the horizontal component?
km
At the instant just before the projectile hits point P, find the horizontal and the vertical components of its velocity (take up and to the right as positive directions).
I got 115.8 m/s (horizontal component)
I got 87.26 m/s (vertical component)-is this right?
What is the the magnitude of the velocity?
m/s. Is magnitude the same as velocity?
What is the angle made by the velocity vector with the horizontal?
° (below the horizontal)


I don't get any of this. I don't know what they mean by range. I don't understand what they are asking me to do.
Can someone explain this to me or show me their work?

2006-10-24 15:35:15 · 2 answers · asked by beast 1 in Science & Mathematics Mathematics

2 answers

The 115.8 and 87.26 are the components at the time of firing (t=0).

The height of the projectile is h = -4.9 t^2 + v0 (t) + h0,

or h = -4.9t^2 + 87.26t + 256

This is 0 when t is approx. 20.27 sec.

The horizontal component is just v0 t. Multiply 20.27 by 115.8 to see how far it goes. That is the range.

2006-10-24 15:53:18 · answer #1 · answered by hayharbr 7 · 0 0

i don't know how far the projectile woould end up above the gun at it's top, but I do know that the projectile will fall this amount as well as the distance of the cliff in precisely in the same time as an object would fall that same distance, it wouldn't matter if it was being shot or not...the time gravity would affect the projectile falling in this distance would be controlled by gravity whether it was shot by a cannon or not....it's hard to jump into this alone, but you need to break it into differnet parts to calc the distance x and y, the time it takes to get to the apogy, and the distance traveled. too much for me this time at night. Vox=118.78, Voy=83.169 m per sec

http://id.mind.net/~zona/mstm/physics/mechanics/curvedMotion/projectileMotion/generalSolution/generalSolution.html

2006-10-24 22:45:56 · answer #2 · answered by Ford Prefect 7 · 0 0

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