Projectile motion problems with cliffs and shot put

  • Thread starter Thread starter ViviCa
  • Start date Start date
Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
3 replies · 4K views
ViviCa
Messages
1
Reaction score
0
hey all!
i've been visiting this forums frm a long time n really the examples here helped me a lot in past.

well now i need some real help!...
im taking AP physics 201 n it Giancoli's 5th Edition.

i need help reagrding projectile motion..?
so I am stating my problem , it seasy tho :frown: but I am not getting them.
i hope n e one can volunteer to help. :smile:

1: A shot-putter throws the shot with an initial speed of 14 m/s at 40 deg angle to the horizontal.Calculate the horizontal distance traveld by the shot if it leaves the athelete's hand at a height of 22.2 m above the ground.

2:A pojectile is shot from the edge of a cliff 125 m above ground level with an initial speed of 105 m/s at an angle of 37.0 deg with the horizontal.(a) dertermine 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.At the instant just before the projectile hits point P, find (c) the horizontal and vertical components of its velocity,(d) the magnitude of the velocity and (e) the angle made by the velocity vector with the horizontal.

3: A ball is thrown horizontally from the top of a cliff with initial speed Vo(at t=0). At any moment, its direction of motion makes an angle (theta) to the horizontal. Derive a formula for (theta) as a function of time,t as the ball follows a projectile's path.

I will highly apppreciate for a help..please!
ViviCa~
 
Physics news on Phys.org
split the problems into componet form
[tex]Vo*cos\alpha[/tex]=x
[tex]Vo*sin\alpha[/tex]=y

then just use your basic kinematic equations
 
hmmm how would u do question 1??

Sin 40= opp/14?
Opposite side = 10 m/sec

So horizontal velocity is 10 m/sec

im lost...