Projectile motion question - please help :)?

Question

a ball is projected off a 100m high cliff at a horizontal speed of 10 m/s.

a) calc. how long the ball takes to hit the sea.
b) calc. horizontal displacement of the ball.
c) calc. the vertical velocity of the ball as it hits the sea.
d) calc. the velocity of the ball as it hits the sea.

2- an archer fires an arrow at a 60m distant target, the arrow leaves the bow at 28m/s at an angle of 25 degrees above the horizontal.

a) what is the horizontal component of the velocity?
b)how long does the arrow take to reach the target ( hits the bulls eye)
c)what is the velocity of the arrow as it hits the target?


3- a golf ball is hit at a velocity of 30m/s at an angle of 30 degrees above the horizontal.

a)calc. max. height of ball
b)the horizontal dist. of the ball.


please help, and please show working

please at least help me do one of these, or show how to do them, like formulae nd stuff,
thanks a lot for any help :) :) +++best answers

Answer 1

It's best to ask one question per post.

1) x(t)=10*t
we need t

vy(t)=-g*t
and
y(t)=100-.5*g*t^2
since we know that when y=0, the ball hits the ground
t=sqrt(100*2/g)
plug this t into x(t) for the distance from the base.

Next, plug the same t into vy(t) to get the vertical component of velocity for that part, and then
v=sqrt(vy^2+100)

2) The horizontal component is
vx(t)=28*cos(25)

since x(t) is 28*cos(25)*t, calculate time to target (I am assuming the target is 60m away horizontally)
60=28*cos(25)*t
t=60/(28*cos(25))
This is the time to target

Now calculate the vertical speed at the time t
vy(t)=28*sin(25)-
g*60/(28*cos(25))

The velocity will be
sqrt(vy^2+vx^2)

now, if the target was exactly 60 m away, but not horizontal, the problem is a bit different
now
60=sqrt(x(t)^2+y(t)^2)
3600=x(t)^2+y(t)^2, solve for t

3600=28^2*cos^2(25)*t^2+
(28*sin(25)*t-.5*g*t^2)^2
Not likely this was intended, but you get the picture

3) x(t)=30*cos(30)*t
vy(t)=30*SIN(30)-g*t
at max height vy(t)=0
t=15/g

This is 50% of the total flight time, so
x(30/g)=30^2*cos(30)/g

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