1. An arrow is moving through xyz-space with the positive z-axis pointing up and
ID: 3345539 • Letter: 1
Question
1. An arrow is moving through xyz-space with the positive z-axis pointing up and with distances measured in meters. The arrow is at the origin at t=0 seconds and its highest point is (250, 0, 490). Give a formula for its position vector at time t assuming that the only force acting on the arrow is the force due to gravity (there is no air resistance).
V_t=
2. A 96-pound object is moving in xyz-space with distances meausred in feet. The total vector force acting on it at time t seconds is F= (3t, -2, 0) pounds, and it at (3,2,1) has a velocity (1, -2, 1) feet per second at t=1. Give formulas for its velocity and position vectors at time t.
r_t=
v_t=
*I'm not sure how to tackle the problem. My first instinct is to use kinematics but appartently it is off of the table according to the professor.*
Explanation / Answer
1. An arrow is moving through xyz-space with the positive z-axis pointing up and with distances measured in meters. The arrow is at the origin at t=0 seconds and its highest point is (250, 0, 490). Give a formula for its position vector at time t assuming that the only force acting on the arrow is the force due to gravity (there is no air resistance).
FORMULA FOR ARROW TRAVEL...
VERTICAL COMPONENT ...OF VELOCITY = VSIN[F] M/SEC
WHERE F IS THE ANGLE OF TRAJECTORY AT WHICH THE ARROW WAS SHOT AT THE BEGINING ..
MAXIMUM HT REACHED = 490-0=490. M ......VERTICAL COMPONENT ...
USING THE FORMULA ...............V^2 - U^2 = 2AS
U = INITIAL VELOCITY = VSIN[T].........
V=FINAL VELOCITY = 0 ...
A=-G=-9.8 M/SEC^2
S= 490 M ..
[ V^2]*[SIN^2(F) ] = 2 * 9.8 * 490..............................................1
TIME OF TRAVEL = T = [V-U]/A = V*SIN(F) / 9.8...SEC.
HORIZONTAL COMPONENT OF VELOCITY = V*COS[F] M/SEC
HORIZONTAL TRAVEL = 250-0=250 M
250 =V*COS(F)*T=[V^2][SIN(F)COS(F)] / 9.8...
[V^2][SIN(F)COS(F)] = 250*9.8...................................................2
EQN.1/ EQN.2 GIVES ......TAN[F]=2*9.8*490 / (250*9.8) ...
.F= 1.321 RADIANS = 75.69 DEGREES ...................................3
SO WE GET ...V= 98/SIN(F) = 101.1385 M/SEC.......................4
SO WE GET ......AT ANY TIME T ......
X = VCOS(F)*T .....Y=0......Z= VSIN(F)*T- 4.9T*T...
WHERE V AND F ARE AS GIVEN ABOVE AND T IS TIME
V_t=[ ( 25T ) , ( 0 ) , ( 98*T - 4.9*T*T ) ]...ANSWER
2. A 96-pound object is moving in xyz-space with distances meausred in feet. The total vector force acting on it at time t seconds is F= (3t, -2, 0) pounds, and it at (3,2,1) has a velocity (1, -2, 1) feet per second at t=1. Give formulas for its velocity and position vectors at time t.
POWER FAILURE SHALL COMPLETE LATE ..UP LOADING FOR THE PRESENT ...
r_t=
v_t=
*I'm not sure how to tackle the problem. My first instinct is to use kinematics but appartently it is off of the table according to the professor.*
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