Use the approximation that for each time step. You throw a metal block of mass 0.27 kg into the air, and it leaves your hand at time 0 at location <0, 2, 0> m with velocity <3.5, 4.0, 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 0.05 s, 0.10 s, and 0.15 s. (Express your answers in vector form.) 0.05 s < , , > m 0.05 s < , , > m/s 0.10 s < , , > m 0.10 s < , , > m/s 0.15 s < , , > m 0.15 s < , , > m/s
An initial position (x0) in vector form which will be given as -
x0 = (2 m)
An intial velocity (v0) in vector form which will be given as -
v0 = (3.5 m/s) + (4 m/s)
(i) The position of a metal block after time (t) which will be given by -
x = x0 + v0 t + (1/2) a t2
At t = 0.05 s, we get
x = (2 m) +
[(3.5 m/s)
+ (4 m/s)
]
(0.05 s) - [(0.5) (9.8 m/s2)
(0.05 s)2]
x = (2 m) +
(0.175 m)
+ (0.2 m)
-
(0.01225 m)
x = (0.175 m) + (2.187 m)
The velocity of a metal block after time (t) which will be given by -
v = v0 + a t
At t = 0.05 s, we get
v = [(3.5 m/s) + (4 m/s)
] -
(9.8 m/s2)
(0.05
s)
v = (3.5 m/s) + (4 m/s)
-
(0.49 m/s)
v = (3.5 m/s) + (3.51
m/s)
(ii) The position of a metal block after time (t) which will be given by -
x = x0 + v0 t + (1/2) a t2
At t = 0.10 s, we get
x = (2 m) +
[(3.5 m/s)
+ (4 m/s)
]
(0.10 s) - [(0.5) (9.8 m/s2)
(0.10 s)2]
x = (2 m) +
(0.35 m)
+ (0.4 m)
-
(0.049 m)
x = (0.35 m) + (2.35 m)
The velocity of a metal block after time (t) which will be given by -
v = v0 + a t
At t = 0.10 s, we get
v = [(3.5 m/s) + (4 m/s)
] -
(9.8 m/s2)
(0.10
s)
v = (3.5 m/s) + (4 m/s)
-
(0.98 m/s)
v = (3.5 m/s) + (3.02
m/s)
(iii) The position of a metal block after time (t) which will be given by -
x = x0 + v0 t + (1/2) a t2
At t = 0.15 s, we get
x = (2 m) +
[(3.5 m/s)
+ (4 m/s)
]
(0.15 s) - [(0.5) (9.8 m/s2)
(0.15 s)2]
x = (2 m) +
(0.525 m)
+ (0.6 m)
-
(0.11025 m)
x = (0.525 m) + (2.489 m)
The velocity of a metal block after time (t) which will be given by -
v = v0 + a t
At t = 0.15 s, we get
v = [(3.5 m/s) + (4 m/s)
] -
(9.8 m/s2)
(0.15
s)
v = (3.5 m/s) + (4 m/s)
-
(1.47 m/s)
v = (3.5 m/s) + (2.53
m/s)
Use the approximation that for each time step. You throw a metal block of mass 0.27...
Use the approximation that for each time step. You throw a metal block of mass 0.24 kg into the air, and it leaves your hand at time 0 at location <0, 2, 0> m with velocity <2.5, 5.5, 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 0.05 s, 0.10 s, and...
Use the approximation thVavg Pflm for each time step You throw a metal block of mass 0.16 kg into the air, and it leaves your hand at time t = 0 at location <0, 2, 0> m with velocity <6.5, 3.0, 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 =...
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,...
Problem 2.21 Use the approximation that Vavg lm for each time ster You throw a metal block of mass 0.22 kg into the air, and it leaves your hand at time 1-0 at location <0, 2, 0> m with velocity <6.0, 7.0, 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...
You throw a metal block of mass 0.22 kg into the air, and it leaves your hand at time t=0 at location < 0, 2, 0 >m with velocity< 2.5, 6.0, 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 = 0.05s, t =0.10s, and t=0.15. (Express your answers in vector form.) x(t...
Please be very specific on
equations used
Use the approximation that Vavg p f/m for each time step You throw a metal block of mass 0.16 kg into the air, and it leaves your hand at time t = 0 at location 0, 2, 0> m with velocity 4.0, 6.0, 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...
Use the approximation that Vavg lm for each time step A spring with a relaxed length of 25 cm and a stiffness of 20 N/m stands vertically on a table. A block of mass 69 g is attached to the top of the spring. You pull the block upward, stretching the spring until its length is now 29.7 cm, hold the block at rest for a moment, and then release it. Using a time step of 0.1 s, predict the...
Use the approximation that for each time step. A spring with a relaxed length of 25 cm and a stiffness of 16 N/m stands vertically on a table. A block of mass 78 g is attached to the top of the spring. You pull the block upward, stretching the spring until its length is now 29.5 cm, hold the block at rest for a moment, and then release it. Using a time step of 0.1 s, predict the position and...
Problem 2.42 (Multistep) Use the approximation that Va for each time step A spring with a relaxed length of 25 cm and a stiffness of 12 N/m stands vertically on a table. A block of mass 67 g is attached to the top of the spring. You pull the block upward, stretching the spring until its length is now 29.1 cm, hold the block at rest for a moment, and then release it. Using a time step of 0.1 s,...
You throw a baseball directly upward at time f = 0 at an initial speed of 12.5 m/s. What is the maximum height the ball reaches above where it leaves your hand? At what times does the ball pass through half the maximum height? Ignore air resistance and take g = 9 80 m/s^2.