A rocket is fired at an angle from the top of a tower of height h_0 = 61.8{\ m m

Question

A rocket is fired at an angle from the top of a tower of height h_0 = 61.8{ m m} . Because of the design of the engines, its position coordinates are of the form x(t) = A + Bt^2 and y(t) = C + Dt^3, where A, B, C , and D are constants. Furthermore, the acceleration of the rocket 1.19{ m s} after firing is
ec a = ( 4.45 hat i+ 1.45 hat j);{ m m/s}^2.
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) = 2Bt{hat i} + 3Dt^2 {hat j} and a(t) = 2B {hat i} + 6Dt {hat j}

a) Find the constants A, B, C, & D

b) At the instant after the rocket is fired, what is its acceleration vector?

c) At the instant after the rocket is fired, what is its velocity?

d) What are the x- and y-components of the rocket's velocity 15.2{ m s} after it is fired?

e) How fast is it moving 15.2{ m s} after it is fired?

f) What is the position vector of the rocket 15.2{ m s} after it is fired?

Explanation / Answer

a)Initial Position (0,61.8) so at time t=0
for x(t)=A+Bt^2 --> 0=A+B(0)2 --> A=0
for y(t)=C+Dt^3 -->61.8=C+D(0)3--> C=61.8

Acceleration equation for x is second derivative for x(t) --> ax=2B,
4.45i =2B --> B=2.225

Acceleration equation for y is second derivative for y(t) --> ay=6Dt
at t=1.19s, 1.45j=6D(1.19s) -->D=0.2

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.