4 home work problems over projectile motion


by man on fire
Tags: home work, motion, problems, projectile
man on fire
man on fire is offline
#1
Nov2-11, 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
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PeterO
PeterO is offline
#2
Nov2-11, 07:08 PM
HW Helper
P: 2,316
Quote Quote by man on fire View Post
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 30.1 i am not sure why i am wrong or if i am wrong)

please explain how you got your answer
You need to show your working for us to comment on what you might be doing wrong.

You have given no indication at all for parts 1 and 2 ??
GR + QM
GR + QM is offline
#3
Nov2-11, 07:10 PM
P: 4
Has your teacher taught you the X and Y table method?

man on fire
man on fire is offline
#4
Nov2-11, 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
zgozvrm
zgozvrm is offline
#5
Nov3-11, 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)
Highway
Highway is offline
#6
Nov3-11, 12:39 PM
P: 349
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
speed is velocity, there should be 4 linear equations of motion that should describe this. . .
PeterO
PeterO is offline
#7
Nov3-11, 07:03 PM
HW Helper
P: 2,316
Quote Quote by man on fire View Post
no she has not taought us the X and Y table method please explain
For projectile motion, the initial velocity can be divided into a vertical component and a horizontal component, using trig. Some people refer to those as an x-component and a y-component as the x axis is traditionally horizontal, while the y-axis is traditionally vertical [when drawn on the page]

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 Vv = 130 and Vh = 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 Vh has a value, and it is 40% of the initial velocity. Trig should give you the answer.


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