Question

A model rocket is launched straight upward with an initial speed of 42.0 m/s. It accelerates with a constant upward acceleration of 2.50 m/s² until its engines stop at an altitude of 160 m.

(a) What can you say about the motion of the rocket after its engines stop?.

(b) What is the maximum height reached by the rocket?

(c) How long after liftoff does the rocket reach its maximum height?

(d) How long is the rocket in the air?

Answer 1

(a)

After its engine stops, rocket behave as a freely falling body with having vertical upward velocity and downward gravitational acceleration. After its maximum altitude , it falls back to the Earth ,gaining speed as it falls.

(b.)

Given at height of 160 m engine stops,

So, velocity at that time is given by kinematics law,

v^2 - u^2 = 2*a*S

here, u = initial velocity = 42.0 m/s

a = 2.50 m/s^2

S = 160 m

v = final velocity = ??

v = sqrt(42.0^2 + 2*2.50*160)

v = 50.6 m/s

now rocket moves with speed 'v' and accleration(-g) = -9.81 m/s^2

also at maximum altitude it's velocity becomes zero

So, by kinematics law,

v^2 - u^2 = 2*a*S

now here, v = 0

u = +50.6 m/s

a = -g = -9.81 m/s^2

S = height gained = ??

S = (0^2 - 50.6^2)/(2*(-9.81))

S = 130 m

So, maximum height reached by rocket = H = 160 + 130

H = 290 m

(c)

total time = (time taken to reach at 160 m (t1)) + (time taken to reach at maximum altitude after engine stops(t2))

for t1,

by kinematics law,

v - u = a*t

here, v = 50.6 m/s

u = 42.0 m/s

a = 2.50 m/s^2

t = t1 = ??

So, t1 = (50.6 - 42.0)/2.50 = 3.44 sec.

for t2,

now,by kinematics law,

v - u = a*t

here, v = 0 m/s

u = 50.6 m/s

a = -9.81 m/s^2

t = t1 = ??

So, t1 = (0 - 50.6)/(-9.81) = 5.16 sec.

therefore total time = T = t1 + t2

T = 3.44 + 5.16

T = 8.6 sec.

(d)

now time taken by rocket to free fall of 290 m is given by,

H = u*t + 0.5*a*t^2

here, H = -290 m

u = velocity at maximum altitude = 0

a = -9.81 m/s^2

t = ??

So, t = sqrt(2*(-290)/(-9.81))

t = 7.69 sec.

So, time of flight = T + t = 8.6 + 7.69

time of flight = 16.3 sec.

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