- #1

man on fire

- 2

- 0

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 into 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.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 into Dy = (v*sin(theta)*t) + .5*g*(t^2) ----> Dy = (61*sin(24)*2.0098) + .5*9.8*2.0098^2} is this correct

Last edited: