4 home work problems over projectile motion

man on fire

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

PeterO

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

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

man on fire

no she has not taought us the X and Y table method please explain

zgozvrm

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

speed is velocity, there should be 4 linear equations of motion that should describe this. . .

PeterO

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 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.