1. A ball is kicked at the edge of 30-m-high cliff with a speed of 15m/s and an angle of 50° to the horizontal as shown. (a) How long will the ball be in the air? (b) How far will the ball land on the ground from the foot of the cliff?
    Solution. 

    (a) We choose the coordinate system with the starting point as the origin and upward direction as +y. 

          We find the components of the initial velocity as

               vx0 = v0cosq0 = (15m/s)cos50° = 9.6m/s;

               vy0 = v0sinq0 = (15m/s)sin50° = 11.5m/s.

         We find the time from the vertical motion

                y = y0 + vy0 t + 1/2gt2;

               -30m = 0 + (-11.5m/s)t + 1/2(9.80m/s2)t2, which gives t = -1.6s, 3.9s.

         We have t = 3.9s.

    (b) We find the range from

               x = vxt = vx0t = (9.64m/s)(3.9s) = 37.6m.

    		  
  2. A 1500-kg sport car was accelerated from rest to 60km/h over a distance of 75m. What was the force the engine exerted on the car?

          Solution. We find the acceleration of the car from the motion equation

                          v2 = v02 + 2 a(x -x0);

                         [(60km/h)(1000m/km)/(3.6ks/h)]2 = 0 + 2a(25m - 0), which gives a = 1.85m/s2.

                         We find the force from

                         F = ma = (1500kg)(1.85m/s2) = 2.8 x 103N. (in the car's moving direction)