A rocket is fired at an angle from the top of a tower of height h0 = 37.8 m . Be
ID: 1414405 • Letter: A
Question
A rocket is fired at an angle from the top of a tower of height h0 = 37.8 m . Because of the design of the engines, its position coordinates are of the form x(t)=A+Bt2 and y(t)=C+Dt3, where A, B, C , and D are constants. Furthermore, the acceleration of the rocket 1.58 s after firing is a? =( 2.60 i^+ 4.55 j^)m/s2. Take the origin of coordinates to be at the base of the tower. Now we can use calculus to determine expressions for the velocity and acceleration of the rocket. v(t)=2Bti^+3Dt2j^ and a(t)=2Bi^+6Dtj^
What are the x- and y-components of the rocket's velocity 19.0 s after it is fired?
How fast is it moving 19.0 s after it is fired?
What is the position vector of the rocket 19.0 s after it is fired?
Explanation / Answer
from given data
B = 2.6
6D -= 4.55/1.58
D = 0.4799
G)
Vx = 2Bt
Vx = 98.8 m/sec
Vy = 519.73 m/sec
H)
v= sqrt (Vx^2 +Vy^2)
v = 529 m/sec
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