Question

Use the approximation that v→avg≈p→f/m for each time step. You throw a metal block of mass 0.34 kg into the air, and it leaves your hand at time t= 0 at location <0, 2, 0> m with velocity <4.5, 3.5, 0> m/s. At this low velocity air resistance is negligible. Using the iterative method shown in Section 2.4 with a time step of 0.05 s, calculate step by step the position and velocity of the block at t= 0.05 s, t= 0.10 s, and t= 0.15 s. (Express your answers in vector form.) r→(t= 0.05 s )= < , , > m v→(t= 0.05 s )= < , , > m/s r→(t= 0.10 s )= < , , > m v→(t= 0.10 s )= < , , > m/s r→(t= 0.15 s )= < , , > m v→(t= 0.15 s )=

Answer 1

Xi = (0 , 2 ,0) m

Vi = (4.5, 3.5 , 0) m/s

a = (0, -9.81 , 0) m/s2 { acc due to gravity)

v(t) = Vi + at

v(0.05) = (4.5, 3.5 , 0) + (0, -9.81*0.05 , 0)

v(0.05) = (4.5 ,3.009 , 0 ) m/s

v(0.10) = (4.5, 3.5 , 0) + (0, -9.81*0.10 , 0)

v(0.10) = (4.5 , 2.51 , 0 ) m/s

v(0.15) = (4.5, 3.5 , 0) + (0, -9.81*0.15 , 0)

v(0.15) = (4.5 , 2.03 , 0 ) m/s

x(t) - xi = vi * t + a*t² /2

x(t) = vi * t + a*t² /2 + xi

x(0.05) = (4.5*0.05, 3.5*0.05 , 0) + (0, -9.81*0.05²/2 , 0) + (0 , 2 ,0)

x(0.05) = ( 0.225, 2.162, 0) m

x(0.10) = (4.5*0.10, 3.5*0.10 , 0) + (0, -9.81*0.10²/2 , 0) + (0 , 2 ,0)

x(0.10) = ( 0.45, 2.301, 0) m

x(0.15) = (4.5*0.15, 3.5*0.15 , 0) + (0, -9.81*0.15²/2 , 0) + (0 , 2 ,0)

x(0.15) = ( 0.375, 2.415, 0) m