Question

A rocket is launched at an angle of 53.0° above the horizontal with an initial speed of 103 m/s. The rocket moves for 3.00 s along its initial line of motion with an acceleration of 28.0 m/s². At this time, its engines fail and the rocket proceeds to move as a projectile.

(a) Find the maximum altitude reached by the rocket. (m)

(b) Find its total time of flight. (s)

(c) Find its horizontal range. (m)

Answer 1

a.)

Using 2nd kinematics law,

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

given, u = 103 m/s

t = time = 3.00 sec.

a = 28.0 m/s^2

So, S = 103*3.00 + 0.5*28.0*3.00^2

S = 435 m

vertical displacement of rocket at end of powered flight(y) = 435*sin(53.0 deg) = 347.41 m

Horizontal displacement of rocket at end of powered flight(x) = 435*cos(53.0 deg) = 261.79 m

now, speed of rocket at end of powered flight,

by 1st kinematics law,

v - u = a*t

v = 103 + 28.0*3.00 = 187 m/s

then, vertical final speed of rocket at end of powered flight(vy) = v*sin(53.0 deg) = 187*sin(53.0 deg) = 149.34 m/s

Horizontal final speed of rocket at end of powered flight(vx) = v*cos(53.0 deg) = 187*cos(53.0 deg) = 112.54 m/s

now using 1st kinematics law in vertical direction,

V - U = a*t

here, V = final speed at maximum height = 0

U = vy = 149.34 m/s

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

t = time = t1 (Let)

So, t1 = (0 - 149.34)/(-9.81)

t1 = 15.22 sec.

So, vertical displacement after end of powered flight,

H = vy*t1 + 0.5*g*t1^2

H = 149.34*15.22 + 0.5*(-9.81)*15.22^2

H = 1.137*10^3 m

So,  maximum altitude reached by the rocket(h) = y + H = 347.41 + 1.137*10^3

h = 1484.41 = 1.484*10^3 m

b.)

now, time taken in fall can be given by,

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

here, t = t2(Let)

u = initial speed at maximum height = 0

a = -9.81 m/s^2

S = -h = -1.484*10^3

So, t2 = sqrt(2*(-1.484*10^3)/(-9.81)) = 17.4 sec.

then, total time of flight = T = t+t1+t2

T = 3.00 + 15.22 + 17.4

T = 35.62 sec.

c.)

total horizontal displacement during free fall can be given by,

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

here, t = t1 + t2 = 15.22 + 17.4 = 32.62 sec.

u = initial speed = vx = 112.54 m/s

a = horizontal acceleration = 0

So, S = 112.54*32.62 = 3671.05 = 3.671*10^3 m

then, horizontal range = R = x + S

R = 261.79 + 3671.05

R = 3.93*10^3 m

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