A rock is thrown horizontally off a 100 m high-rise building as shown in Figure Q4(a). It lands 95 m away. At what speed was it thrown

P3.8 For the mechanism shown in Figure P3.8, deter- O mine the linear velocity of point D using instanta- neous centers. = 2.5 in; ro.A = 1.0 in =D1.5 in; fo,c = 1.0 in; fo,0. "o,0, TAB = 2.5 in; ro,B =1.5 in; ro.c = 2.0 in Tcp = 1.0 in; rcE = 1.5 in; ro.E 0, = 45°; 0, = 5.0 rad/sec CCW %3D %3D = 2.0 in %3D %3D 3 4. PA 2. 02 O, O 1 1 9. DE Figure P3.8

In a nutshell, a smaller rock is propelled into the air as a huge counterweight in the back falls down, effectively converting gravitational potential energy into kinetic energy for the rock and the weight. But, since the rock is smaller.. it is propelled at a much faster velocity! Let's say we wanted to hurl a 50 kg boulder at a castle wall 100 meters away. Our wooden structure for the trebuchet is 10 meters tall, so that is the maximum distance the counterweight can fall. Assuming the launch angle is a perfect 45 degrees, and the boulder is released from ground level, derive the needed mass of the counterweight that will give the boulder enough kinetic energy to reach the base of the castle walls. You may assume no energy is "lost" due to friction, and ALL the counterweight's gravitational potential energy is converted into the boulder's kinetic enerqy. (HINT: You will need to employ 2D kinematics to find the needed final velocity, then use that final velocity in an energy equation…

VP5.22.3 An aerobatic airplane flying at a constant 80.0 m/s makes a horizontal turn of radius 175 m. The pilot has mass 80.0 kg. (a) What is the bank angle of the airplane? (b) What is the pilot’s appar- ent weight during the turn? How many times greater than his actual weight is this?