Question

Chapter D2, Problem D2/134 A projectile is launched from point O with an initial speed Vo 515 ft/sec directed as shown in the figure. Compute the x-, y and z-components of position, velocity, and acceleration 22.5 seconds after launch. Neglect aerodynamic drag 63 ok 20 Answers: At t 22.5 sec, х3 ft ft/sec, ft/sec2 у 3 t/sec f/sec, ay ft, az= ft/sec, ft/sec2

Answer 1

Here, we consider one axis i.e.X'(by replacing X and Y axis) at an angle 20^o from X axis, and after that we will find out position, velocity and acceleration into X and Y direction at 22.5 second by resolving position, velocity and acceleration into X' direction at 22.5 second.V(as per following below).

Position:

X= X'cos20^o__________________(1)

Y= X'sin20^o__________________(2)

Velocity:

V_X= V_X'cos20^o_________________(3)

V_Y= V_X'sin20^o__________________(4)

Acceleration:

A_X= A_X'cos20^o __________________(5)

A_Y= A_X'sin20^o___________________(6)

Vo: 615ft/Rec--- 1 -1

Step-1 To find out acceleration into X' and Z directions

Into X' direction:

as per mention into question, 'to neglect aerodynamic drag' it means there is no acceleration into X' direction

a_X'=0 ft/s²

Now put above value into equation (5) and (6)

a_X = 0cos20^o = 0 ft/s²

a_Y = 0sin20^o = 0 ft/s²

Into Z direction:

as we know, into vertical direction, there is always gravitational acceleration.

a_Z = -g = -32.17 ft/s²

Step-2: To find out velocity into X' and Z direction.

Into X' direction:

V_X' = V_ocos63^o - a_xt = 515cos63^o - 0= 233.80 ft/s

Now put above values into equations (3) and (4)

V_X = V_X'cos20^o = 233.80cos20^o = 219.70 ft/s

V_Y = V_X'sin20^o = 233.80sin20^o = 79.96 ft/s

Into Z direction:

V_Z = V_osin63^o - gt= 515sin63^o - 32.17(22.5) = 264.96 ft/s

Step-3: To find out displacement and position into X' and Z directions

Into X' direction:

X' = V_X'.t -0.5a_X't²= 233.80(22.5) - 0 = 5260.5 ft

put above values into equations (1) and (2)

X = X'cos20^o = 5260.5cos20^o = 4943.25 ft

Y = X'sin20^o = 5260.5sin20^o = 1799.19 ft

Into Z direction:

Z = V_osin63^ot -0.5gt² = (22.5)515sin63^o - 0.5(32.17(22.5²)) = 2181.50 ft

Thus, components of position, velocity and acceleration in X, Y and Z direction are followed:

	Positions(ft)	Velocity(ft/s)	Acceleration(ft/s2)
X- Direction	4943.25	219.70	0
Y- Direction	1799.19	79.96	0
Z- Direction	2181.50	264.96	-32.17