Denote the initial speed of a cannon ball fired from a battleship

[nrc_8706]

denote the initial speed of a cannon ball fired from a battleship as Vo. when the initial projectile angle is 45 degrees with respect to the horizontal, it gives a maximum range of R.
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)

 

[VietDao29]

So, what have you done? Which option to you is correct?
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,

 

[nrc_8706]

thanks

thank you viet dao, i figured it out.

 

[nrc_8706]

?

but now how do you find max height?
do u use y=voyt + 1/2gt^2?

 

[VietDao29]

The y-component of the velocity makes the object go up or down.
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 0 (m / s). The object's at its max height, because, right after that, it will start accelerate downwards (ie, it no longer moves upwards).
You can use:
v_f² = v_i² + 2ad
Here a = -g (if you choose the positive direction upward).
v_i is the object's initial velocity.
v_f is the object's final velocity.
Here, you just need the y-component, so the v_i is the y-component of the initial velocity.
v_f = 0 m / s. It's when the object's velocity has no more y-component.
You can use that and solve for d, which's the object's max height.
Viet Dao,