draw or make an illustration to this problem 1 and 2

1) Projectile motion problem : A ball is launched with a speed of 5 m /s at an angle 60 above the horizontal. Find its velocity at the highest point of trajectory and the horizontal range of the projectile Solution : A) The velocity at the highest point of trajectory is the horizontal component of the initial velocity as the vertical component of the velocity vanishes at the highest point of trajectory velocity at highest point = 5 Cos60 m/s = 2.5 m/s B) The horizontal range is given by the formula : Range = v2sin (20) 9 Range = (5)2Sin (120) 9.8 Range = 2.21 m 2 ) Equilibrium model A meter stick is balanced by a fulcrum placed at the 50 cm mark. a weight of 10 N is suspended at 20 cm mark, find the weight that must be suspended at 90 cm mark , so as to balance the meterstick at the fulcrum (pivot) Solution : Using the condition of rotational equilibrium (balancing of torque) Torque due to weight on left = torque due to weight on right 10 N x (50 cm - 20 cm) = wx (90 cm - 50 cm) 10 N X 30 cm = w x 40 cm 7.5 N = W Therefore, a weight of 7.5 N is must be suspended at 90 cm mark