I'm having trouble with these 2 questions. I've tried them and come up with solutions but would like someone to check if I have done them properly. A traditional watch has a second hand 1.5 cm long from the center to the tip. a) What is the speed of the tip of the second hand? d = 2Pir d = 2(3.14)(1.5) v = d/t v = 2(3.14)(1.5)/60s v = 0.157 cm/s b) what is the velocity of the tip at 15s? 45s? 60? 0.157 cm/s (right) 0.157 cm/s (left) 0.157 cm/s (up) c) What is the change in velocity between 30s and 45s? I'm having trouble with this one. Do I use the pythagorean theorem because one is pointing to the left and one is pointing down. 0.157^2 + 0.157^2 - sq root v = 0.222 cm/s (sw) A student tosses their pen vertically upward with an intial velocity of 3.8 m/s. a) Maximum height Vf^2 = Vi^2 + 2ad 0 = 3.8^2 + 2(-9.8)d d = 0.736735 m b) How much time will pass before the pen returns to his hand if it is at the same level that he released the pen at? d = v2t - 1/2at^2 0= (3.8)(t) - 1/2(9.8)t^2 t = 0.775510 s c) if the student's hand is 1.5 m above the ground and he misses the pen, with what speed will the pen hit the ground. 0.736735 + 1.5 = 2.236735 m Vf^2 = Vi^2 + 2ad Vf^2 = (0) + 2(9.8)(2.236735) vf = 6.62117 m/s (down) Can someone plz check these over. I would really appreciate it. Thank you.