Afaulty model rocket moves in the xy-plane (the positive y direction is vertically upward). the rocket’s acceleration has components ax(t) = αt2 and ay(t) = β −γt, where α = 2.50 m/s4 , β = 9.00 m/s2 , and γ = 1.40 m/s3 . at t = 0 the rocket is at the origin and has velocity ~v0 = v0x ˆi + v0y ˆj with v0x = 1.00 m/s and v0y = 7.00 m/s. (a) calculate the velocity and position vectors as functions of time. (b) what is the maximum height reached by the rocket? (c) sketch the path of the rocket. (d) what is the horizontal displacement of the rocket when it returns to y = 0?

part a)

velocity is given as

position is given as

Part b)

maximum height is given as

Part c)

Path will be the curved path

Part d)

displacement in x direction will be

Explanation:

also we know the initial velocity as

part a)

from acceleration in x direction we can say

now again integrate both sides

now similarly for y direction

now again integrate both sides

PART B)

For maximum height we know that velocity in y direction must be zero

so we will have

by solving above we have

now at this time the height is given as

PART C)

path of the rocket will be curved path in xy plane

PART D)

when y = 0 then we have

by solving above equation we have

now at this time position of X is given as

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its a 5.5 n

glad that