Hi all, I asked this question last week butI still need some more help with it.

Question

Hi all, I asked this question last week butI still need some more help with it. I have answered the majority of this problem, butI dont know if ive done it correctly. I have used standard metricsystem formulae to solve for the projectiles' flight path,but I need to know how to include the ground surface profileinto my calculations for a projectiles' flight path. I havesketched the profile using a graphing calculator, but would like toknow how to solve this manually.
At t=0, a 2 kilogram projectile is located at the origin and has avelocity of 22 metres/second at an angle of 40 degrees above thehorizontal. The profile of the ground surface can be approximatedby the equation y= 0.6x –0.005x^2where xand y are in meters. Sketch theprofile of the ground surface and the path taken by theprojectile and determine the approximate coordinates ofthe point where the projectile hits the groundh. Any help would be mostappreciated. -Nick. Hi all, I asked this question last week butI still need some more help with it. I have answered the majority of this problem, butI dont know if ive done it correctly. I have used standard metricsystem formulae to solve for the projectiles' flight path,but I need to know how to include the ground surface profileinto my calculations for a projectiles' flight path. I havesketched the profile using a graphing calculator, but would like toknow how to solve this manually.
At t=0, a 2 kilogram projectile is located at the origin and has avelocity of 22 metres/second at an angle of 40 degrees above thehorizontal. The profile of the ground surface can be approximatedby the equation y= 0.6x –0.005x^2where xand y are in meters. Sketch theprofile of the ground surface and the path taken by theprojectile and determine the approximate coordinates ofthe point where the projectile hits the groundh. Any help would be mostappreciated. -Nick.

Explanation / Answer

initial velocity is V=22 m/s at 40 degrees to horizontal(x-axis). x-component = u = V cos = 22 cos40 = 16.85 m/s y-component = v = V sin = 22 sin40 = 14.14 m/s in time t, the vertical displacement y & horizontaldisplacement x are y = vt - (1/2)gt^2 x = ut so we express y in terms of x as y = vt - (1/2)gt^2 = v(x/u) - (1/2)g(x/u)^2 y = .84x - .017x^2 >>> parabola describing the path of theprojectile this intersects the ground profile y = .6x - .005x^2 at (X,Y)where .84X - .017X^2=.6X - .005X^2 .24X=.012X^2 X = .24/.012 = 20 m Y = .6*20-.005*400= 10 m

Related Questions

Navigate

• Browse (All)
• Browse H
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.