Physics - projectile motion question.?

Question

The range of a cannonball fired horizontally from a cliff is equal to the height of the cliff. What is the direction of the velocity vector when the projectile strikes the ground?
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Answer 1

PAY ATTENTION........

at the monment the cannonball hits the ground you have two perpendicular vector summing into a total velocity vector with a certain direction. if you make he diagram you eill end up that

arctaN (Ux / Uy) = angle.


now you know that in horizontal axis yhe cannon ball moves at a constant speed therefore

Ux = S / t => S= Ux * t

and in the vertical axis it moves under the acceleration of gravity;


so height (H) = 1/2 g* t^2 = 1/2 gt * t ,,,now
Uy = gt therefor,

H = 1/2 Uy * t

we know that H=S therefore

we can to prove that 1/2 Uy*t = Ux * t ===>>>>


Ux/ Uy = 1/2

so we return to the arctan and say that arctan (1/2) means that the angle is 26.56degres


right ' ;)

or to be more correct if zero degrees is in the same direction of Ux then theta = 295.26

Answer 2

a cannonball shot horizontally from a very high cliff at a high speed. And suppose for a moment that the gravity switch could be "turned off" such that the cannonball would travel in the absence of gravity? What would the motion of such a cannonball be like? How could its motion be described? According to Newton's first law of motion, such a cannonball would continue in motion in a straight line at constant speed. In the absence of all forces, "an object in motion will ...". This is Newton's law of inertia.








Now suppose that the "gravity switch is turned on" and that the cannonball is projected horizontally from the top of the cliff. What effect will gravity have upon the motion of the cannonball? Will gravity effect the cannonball's horizontal motion? Will the cannonball travel a greater (or shorter) horizontal distance due to the influence of gravity? The answer to both of these questions is "No!" Gravity will act downwards upon the cannonball to effect its vertical motion. Gravity causes a vertical acceleration, causing the ball to drop vertically below its otherwise straight-line, inertial path. Gravity is the downward force upon a projectile which influences its vertical motion and causes the parabolic trajectory which is characteristic of all projectiles.

Answer 3

The height of the cliff and the range are equal, and assuming air resistance negligable...

So the time of flight is equal for both the horizontal and vertical componants.

So muzzle velocity (Vm) x T = average falling velocity (Vf ave) * T

Divide both sides by T and you get Vm=Vf ave

but as it strikes the gound, Vf instant will be twice Vf ave

so the vertical componant will be 2 x Vm

So you have to solve the vector diagram where the vertical componant is twice the horizontal one. The acute angle is your answer.

Answer 4

If the piece separates gently (and not with an explosion as in case of a hearth cracker rocket ) then it is going to fly edge by using edge as no exterior rigidity has acted on the stone or the piece. So the desirable answer is selection a million. In case of an explosion halfway it is going to count near to separation. The small piece can bypass in any random direction (not inevitably horizontal) and stick to diverse parabolic course.

Answer 5

h= height of cliff=range. let t be the time of flight. so taking vertical component of velocity =gt. the horizontal distance h=vt or v=h/t the horizontal component of velocity.
so tantheta= gt/h/t=gt^2/h. so now take tan inverse to get the angle.

Answer 6

45 degrees downwards.