Question

A rocket is launched at an angle of θ = 51° above the horizontal with an initial speed vi = 55 m/s, as shown below. It moves for 25 s along its initial line of motion wth an acceleration of 20.4 m/s2. At this time, its engines fail and the rocket proceeds to move as a free body.

(a) What is the rocket's maximum altitude? m

(b) What is the rocket's total time of flight? s

(c) What is the rocket's horizontal range? m

Answer 1

(a)

total distance covered by the rocket in first 23 s is ,

s = u*t + 1/2*a*t²

s = 55*25 + 0.5*20.4*(25)²

s = 7750 m

The height using distance and angle is ,

H = s*sin51 = 7750*sin51

H = 6022.88 m

the horizontl distance for this time is ,

x = s*cos51 =7750*cos51

x = 4877.23 m

final velocity after travelling 25 s,

v = u+at

v = 55 + 20.4*25

v = 565 m/s

The maximaum height is

Hmax = H + (vsin51)²/2g = 6022.88 + (565*sin51)²/2*9.8

Hmax = 6116.09 m

(b)

The time of the flight to reach the maximum height ,

t1 = 25 + 565*sin51/9.8

t1 = 69.80 s

the time of flight back to the ground is,

t2 = sqrt(2*Hmax/g) = sqrt(2*6116.09 /9.8)

t2 = 35.32 s

The time of flight of

t = t1+ t2 = 69.80+35.32

t = 105.12s

(c)

The Horizontal range of ,

R = x + (565*cos51)*(105.12 -25) = 4877.23 +(565*cos51)*(105.12 -25)

R = 33364.69 m