1) (10 pts)A ball is thrown vertically up by a juggler and caught back 
   2 seconds later.
   What is the maximum height reached by the ball?


    Time of flight up = time of flight down = 1s. 
    Consider descending path:
    With the initial velocity at the apogeum v_0=0 the formula for the distance
    reduces to y=1/2 * g * t^2 = 1/2 * 9.8 m/s^2 * 1 s^s = 4.9 m

2) (10 pts)You are on the edge of a vertical cliff that is 19.6 m high 
   and you throw a stone horizontally with a speed of 20 m/s. How far from 
   the foot of the cliff the stone will hit the ground ?

   The time of flight is controlled by vertical drop. Only gravity affects the
   descend of the stone. The initial velocity in vertical direction is zero.
   The relation between height and time is y = 1/2 * g * t^2 .
   This gives us time as t = sqrt (2*y/g) = 2 s.
   The horizontal path for constant velocity is 
     x= v * t = 20 m/s * 2 s = 40 m

3) (20 pts) A motor boat can travel at 10 km/h in still water. A river flows 
   at a speed of 5 km/h west. A boater wishes to cross from the south bank 
   to a point directly opposite on the north bank. At what angle must the boat 
   be headed?

   The angle must be such that the east-west component of the boat velocity
   will cancel the speed of water flow.
    10 km/h * sin(theta) = 5 km/h ----> theta =30 degrees


4) A 1000 kg car is pushed  along a level road by four students who apply a
   total forward force of 100 N. Neglecting friction,
  a) (5 pts)what is the acceleration of the car? 
  b) (5 pts)what speed will the car have after half minute of effort?

     F = m * a  --> a = F/m = 100 (kg m/s^2)/1000 kg = 0.1 m/s^2
     v = a * t = 0.1 m/s^2 * 30 s = 3 m/s

5) a) (5 pts)A man weighing 60 kg is in an elevator that is accelerating upward 
      at 5 m/s^2. What is the total force exerted on him by the elevator floor?
   b) (5 pts)What is the force during the descent with the same value of 
      acceleration?

    F = m * g + m * a = 60 kg * (9.8+5.0)m/s^2 = 888 N
    F = m * g - m * a = 60 kg * (9.8-5.0)m/s^2 = 288 N

6) (20 pts) A tennis ball dropped in a free fall from a hight of 5 m (first
   floor of Dallas Hall) bounces up loosing 30% of its mechanical energy at
   each bounce. You can see the bounce only if the ball moves up by more than 
   11 cm after the bounce. How many bounces should you see?

    Potential energy U = m*g*h
    After the collision U1 = m*g*h1 = 70% U  ---> h1 = 70% h
    height after the bounce :  1 bounce  h1 = 3.5  m
                               2 bounce  h2 = 2.45 m
                               3 bounce  h3 = 1.72 m
                               4 bounce  h4 = 1.20 m
                               5 bounce  h5 = 0.84 m
                               6 bounce  h6 = 0.59 m
                               7 bounce  h7 = 0.41 m
                               8 bounce  h8 = 0.29 m
                               9 bounce  h9 = 0.20 m
                              10 bounce  h10= 0.14 m   --the answer
                              11 bounce  h11= 0.10 m

7) (20 pts) A movie actor has to perform a stunt of driving a motocycle
   onto a a ramp inclined at 30 degrees with respect to the road and jumping 
   from the end of this ramp over two police cars. He will then land on 
   the embankment that is at the same level as the end of the ramp. The total 
   length of the jump must be 10 m in order to reach the embankment. Neglecting
   air resistance and friction what is the minimum speed that the stuntman 
   has to reach on the ramp in order to succeed?

   The range R = v^2*sin(2*theta)/g  --> v=sqrt(R*g/sin(2*theta))
                                          =sqrt(10m*9.8m/s^2/sin(60))
                                          =10.64 m/s = 38.3 km/h