A projectile is fired with initial velocity of vo feet per second. The projectil

Question

A projectile is fired with initial velocity of vo feet per second. The projectile can be pictured as being fired from the origin into the first quadrant, making an angle ? with the positive x-axis. If there is no air resistance, then at t seconds the coordinates of the projectile (in feet) are x=votcos? and y=-16t2+votsin?. Suppose a projectile leaves the gun at 100 ft/sec and ?=60 degrees. a) What are the coordinates of the projectile at time t=4 sec? (200, 90.4) b) For how many seconds is the projectile in the air? c) How far from the gun does the projectile land? d) What is the maximum height attained by the projectile? e)Find an expression in terms of vo and ? for the time in the air. f) Find an expression in terms of vo and ? for the distance from the gun.

g) Find an expression in terms of vo and ? for the maximum height.

Explanation / Answer

your answer to a is correct. (200, 90.4) Im guessing ur -16 is -1/2(gravity), because that's how the real equation should be....and gravity should be 9.8 instead of 8...but o well. B. you just use the equation for y and make y=0 and solve for t. so y=0=-16t^2+100tSin (60) t=5.41 second. C. you plug in t from B, and you get x=5.41(100)Cos(60)=270.5 D An easy way to do this is since there is no air resistence, you can just divide ur T=5.41 by two, cuz the maximum height is always half way between the start and the end. I use calculus to prove this, but idk if u know how to do it, but here goes. find the first derrivative of the y equation, u get y'=-32t+100sin 60, and to find the max point, u find the zero of the derrivative, which is when t=2.7. And you just plug 2.7 into ur y function and u get 117.19. E. Look back to how u do B....notice that u make teh equation y=0. Well this time, instead of pluging in Vo and ?, you leave it as a variable. Then solve for t. so u get y=0=-16t^2+VoTsin(?) solve for t by moving -16t^2 to the other side. Then cancel out 1 t on both side, then divide 16 on both side. u get t=(voSin?)/16 F So u just plug in ur answer from E to ur function x=VotCos? u get x=(Vo^2sin?cos?)/16 Usually, to make it simple, i would simplify it, by making sin?cos?=(1/2)Sin2? That's the trig identity. so u get x=(Vo^2sin2?)/32. G. Like i said before, ur t value where y is at max, is when t=half of the total travel time. And you already found the total travel time equation which is t=(voSin?)/16, so u divide that by two u get (vosin?)/32. And then plug that into ur y function, u get -16((vosin?)/32)^2+vo((vosin?)/32)Sin? you can just simplify it anyway u want i guess. H. so this is basically proving y=-16t^2+votsin? take sec^2? down by making it 1/cos^2 ? and make x^2=(vo^2)(t^2)(cos?)^2 So u cancel out teh (vo^2)*(cos?^2) on the bottom so u get -16t^2+votcos?tan? replace tan? with (sin?)/(cos?), so u can cancel out the cos? u get -16t^2+votsin?=y

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