projectile motion problems (1 Viewer)

[ssssonicyouth]

hi, could someone help me with these ones from fitzpatrick?

17. A stone is thrown so that it will hit a bird at teh top of a pole. However, at the instant the stone is thrown, the bird flies away in a horizontal straight line at a speed on 10m/s. The stone reaches a height double that of the pole, and in its descent, touches the bird. Find the horizontal component of the velocity of the stone.

Answer: 12.1 m/s

20. A particle projected from a point meets the horizontal plane through the point of projection after travelling a horizontal distance 'a', and in the course of its trajectory attains a greatest height 'b' above the point of projection. Find that horizontal and vertical components of the velocity in terms of 'a' and 'b'. Show that when it has described a horizontal distance x, it has attained a height of 4bx(a-x)/a^2

Answer: a/2.root(g/2b) ; root(2gb)

24. Find the speed and direction of a particle which, when projected from a point 15m above the horizontal ground, just clears the top of a wall 26.25m high and 30m away.

Answer: 25 m/s; 36 degrees and 52 minutes

I would feel much better going into the exam understanding them

 

[redslert]

[DELETED]

i got it wrong...
hmm

let me think about it first..

ill get back to you

 

[Fosweb]

For the first one, search the 4U forum, I remember that came up a while ago.

No time to look at the other ones yet though.

 

[jogloran]

The first one is also identical to the Tip for projectile motion with moving target in that 50 Exam Tips book.

 

[TimTheTutor]

Here's how to do Q2

Let V = projected initial velocity, @ = projected initial angle
Therefore Vy (vertical velocity component) = Vsin@
Therefore Vx (horizontal velocity component) = Vcos@

Derive the following two standard results to begin with:

(1) Max Height Reached (b) = V^2sin^2@/2g
(2) Range (a) = 2(V^2)sin@cos@/g

Rearranging (1) we have:
2gb = (Vsin@)^2
Vsin@ = sqrt(2gb)
Vy= sqrt(2gb)

Rearranging (2) we have:
ag/2 = Vsin@Vcos@
Vcos@ = ag/2*1/sqrt(2gb)
Vx = a*sqrt(g^2)/2*sqrt(2gb)
= a/2*sqrt(g^2/2gb)
= a/2*sqrt(g/2b)

For the final part use the cartesian form of the projectile path

y= -gx^2/2(Vcos@)^2 + xtan @
= -gx^2/2(a^2*g)/(2^2*2b) + xtan@
= -4bx^2/a^2 + xtan@

Now tan @ = Vy/Vx = a/2*sqrt(g/2b) / sqrt(2gb) = 4b/a

Therefore y = -4bx^2/a^2 + 4bx/a = 4bx(a-x) / a^2