- #1

nrc_8706

- 71

- 0

the time of flight of the cannonball for this maximum range R is given by

1.t=(1/3^1/2)vo/g

2.t=3^1/2(vo/g)

3.t=2^1/2(vo/g)

4.t=2(vo/g)

5.t=1/2(vo/g)

6.1/(2^1/2)(vo/g)

7.t=4(vo/g)

8.t=(1/4)(vo/g)

9.t=(2/3)(vo/g)

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- Thread starter nrc_8706
- Start date

- #1

nrc_8706

- 71

- 0

the time of flight of the cannonball for this maximum range R is given by

1.t=(1/3^1/2)vo/g

2.t=3^1/2(vo/g)

3.t=2^1/2(vo/g)

4.t=2(vo/g)

5.t=1/2(vo/g)

6.1/(2^1/2)(vo/g)

7.t=4(vo/g)

8.t=(1/4)(vo/g)

9.t=(2/3)(vo/g)

- #2

VietDao29

Homework Helper

- 1,426

- 3

Because you just need to find the time of flight, you don't need the x-component of your initial velocity, you just need the y-component of the initial velocity. So first, you can try to find out the y-component of the velocity.

Then, note that the object has the acceleration of g (downard). Can you find out the time in flight of the object?

Viet Dao,

- #3

nrc_8706

- 71

- 0

thank you viet dao, i figured it out.

- #4

nrc_8706

- 71

- 0

but now how do you find max height?

do u use y=voyt + 1/2gt^2?

- #5

VietDao29

Homework Helper

- 1,426

- 3

Because the object has the acceleration of g (downward), so the y-component of the initial velocity decreases, and the object moves upwards slowlier and slowlier, finally when the y-component of the velocity is

You can use:

v

Here a = -g (if you choose the positive direction upward).

v

v

Here, you just need the y-component, so the v

v

You can use that and solve for d, which's the object's max height.

Viet Dao,

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