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4 home work problems over projectile motion 
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#1
Nov211, 06:26 PM

P: 2

1. a projectile is fired into the air at a projection angle of 30° above the ground. its initial speed is 327 m/s. what is its speed 3 seconds later
2.the speed of a projectile at its maximum height was 40% of its speed at ground level. find the angle of elevation. 3.a man stands on a cliff 15m abouve a level plain. he tosses a rock at 36m/s into the air over the plain. at what angle to the horizontal did he throw the rock if it is in the air for 6.6s(i got 63.93° but my teacher got 56.6° i am not sure why i am wrong or if i am wrong) 4.a stone is projected from ground level aith a speed of 61m/s at an angle of 24° with the horizontal. how high above the ground, assumed level, will the stone be when its horizontal displacement is 122m(i got 69.65m my teahcer got 30.1m i am not sure why i am wrong or if i am wrong) please explain how you got your answer 1. i have no clue what to do 2. still no clue 3. i used {2(sin(theta)*V)/g = t > (2*sin(theta)*36)/9.8 = 6.6 > (6.6*9.8)/(2*36)=sin(theta) > sin^1(64.68/72) = theta} is this correct 4. i used {t = Dh/Vcos(theta) > t = 112/(61*cos(24) > t = 2.0098 > now with time i put it in to Dy = (v*sin(theta)*t) + .5*g*(t^2) > Dy = (61*sin(24)*2.0098) + .5*9.8*2.0098^2} is this correct 


#2
Nov211, 07:08 PM

HW Helper
P: 2,318

You have given no indication at all for parts 1 and 2 ?? 


#3
Nov211, 07:10 PM

P: 4

Has your teacher taught you the X and Y table method?



#4
Nov211, 07:23 PM

P: 2

4 home work problems over projectile motion
no she has not taought us the X and Y table method please explain



#5
Nov311, 12:27 PM

P: 754

For part 3, your teacher is correct.
For part 4, I got 30.8m, assuming g = 9.8 m/s/s (30.4m, if using g = 10 m/s/s) 


#6
Nov311, 12:39 PM

P: 349




#7
Nov311, 07:03 PM

HW Helper
P: 2,318

During the flight, the horizontal component remains constant [until the projectile hits something] and the vertical component follows the normal laws of vertical motion under g. approximation example:  using g=10 for simplicity. Your 327 m/s at 30 degrees amounts to vertical approx 160, horizontal approx 280. In 3 seconds, the vertical component will reduce by 30 [remember I am taking g = 10 for simplicity here] so now V_{v} = 130 and V_{h} = 280 [unchanged]. You can use Pythagoras to reconstitute the speed that means. That's how #1 is done  but with far more accurate figures! #2 can be analysed as above, and at maximum height, the Vertical component = 0, so only V_{h} has a value, and it is 40% of the initial velocity. Trig should give you the answer. 


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