Projectiles help!!!!!!!!! plzzzzzzz?

Question

A projectile is shot from the edge of a cliff h = 225 m above ground level with an initial speed of v0 = 105 m/s at an angle of 37.0° with the horizontal, as shown in Fig. 3-39.

I uploaded Figure 3-39 at this site: http://i20.tinypic.com/16c9oio.jpg

(a) Determine the time taken by the projectile to hit point P at ground level.
(b) Determine the range X of the projectile as measured from the base of the cliff.
(c) At the instant just before the projectile hits point P, find the horizontal and the vertical components of its velocity (take up and to the right as positive directions).
(horizontal component)
(vertical component)
(d) What is the the magnitude of the velocity?
(e) What is the angle made by the velocity vector with the horizontal (below the horizontal)?

Thanks soo much!!!!!!!!!!!!

plz don't post rude remarks. Obviously I have already attempted the problem, and that's why I am posting it here...I do not know how to do this. If you can help, please help. I would realyl really appreciate that. If you are only on here to post rude comments and get 2 points for answering a question, please do not post an answer here.

Answer 1

ok, first thing - we are ignoring air friction - this really does matter and it's always worth telling an examiner that you know this.

the vertical component is v sin(theta), the horizontal component is v cos(theta) - so you should be able to get the vertical velocity.

it will come back past you at the same speed (opposite velocity) so v = -u on the way down. this means that you can use -u_vertical =u_vertical - gt and solve for t.... this brings it back to where you started.

then it continues to fall for another 225 meters... so you can apply v^2=u^2+2g*h and get the final speed. then it's v=u+gt again to get the time difference _after_it_passes_you_on_the_way_down.

(actually, if your math is pretty good, you can use s=ut+1/2 at^2 too... maybe you can try that as an exercise)

b. the range - you know the time the projectile flies for (part a.)and the horizontal speed from the introduction... so this should be easy.

c. ok, so the projectile is going at the same speed horizontally and you've worked out v in part a.

d. it's just about applying pythagoras' theorem

e. the impact angle = atan(vertical velocity / horizontal velocity)

Answer 2

Treat horizontal and vertical velocity components separately. The horizontal component vx is v0*cosø, and that velocity stays the same as long as the projectile is in the air.

The vertical component of initial velocity is v0*sinø. The projectile's vertical velocity will decrease as it climbs because of the acceleration of gravity, 9.8 m/sec^2, so vy is given by

vy = v0*sinø - g*t

It will continue to climb until the vertical velocity is zero. This allows you to calculate the ascent time ta. Using this ta you can then calculate how high the projectile gets above the launch point, since vertical distance ym = v0*ta - 0.5*g*ta^2

Then you can calculate the descent time td; The projectile falls from an altitude ym + h to reach the ground. The descent time is then given by (ym + h) = 0.5*g*td^2.

The total time of flight is then ta + td. The horizontal distance traveled in that time is the range X. Use the horizontal velocity component in the formula distance = velocity x time.

The vertical component of velocity when it hits is g*td (acceleration x descent time). The magnitude of the impact velocity is the square root of the sum of the squares of horizontal and vertical components, and the angle is arctan(vertical vel / horiz. vel).

Answer 3

to start with find the x and y componets of the 37.0 degrees
that is the first step

Use sin and cos to make the 37 degrees and 105 m/s into vertical and horizontal


Ok, i could show you the work, but my computer wont send it


sorry


but it looks like the other people have good answers

Answer 4

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Answer 5

is this a webassign problem?
do your own homework.
Show alittle effort and people will be more like to respond.
It is a vector problem