Question

A rocket is launched at an angle of 39◦ above the horizontal with an initial speed of 90 m/s. It moves for 7 s along its initial line of motion with an acceleration of 19 m/s 2 . At this time its engines fail and the rocket proceeds to move as a projectile. The acceleration of gravity is 9.8 m/s 2 . Find the maximum altitude reached by the rocket. Answer in units of m.

Answer 1

Answer:

Y=1370.23m

Explanation:

The motion have two moments the first one the time the initial velocity is accelerating then when the engines proceeds to move as a projectile

$a=19 \frac{m}{s^{2} } \\voy=vo*sin(\alpha )\\voy=90*sin(39 )\\y_{o}=0m\\y_{f}=y_{o}+v_{oy}*t+\frac{1}{2}*a*t^{2}\\y_{f}=90*sin(39)*7s+\frac{1}{2}*19\frac{m}{s^{2} }*(7)^{2}\\y_{f}=861.97m$

Now the motion the rocket moves as a projectile so:

$v_{fy}=v_{iy}+a*t\\v_{fy}=90+9.8*7\\v_{fy}=158.6 sin(39)$

Now the final velocity is the initial in the second one

$v_{fy}^{2}=v_{fi}^{2}+2*a*yf \\\\a=g$

The maximum altitude Vf=0

$0=v_{fi}^{2}+2*a*yf \\\\yf=\frac{(158.6 sin(39))^{2} }{2*9.8\frac{m}{s^{2} } } \\yf=508.26m$

So total altitude is both altitude of the motion so:

$Y=508.2m+861.97m\\Y=1370.23m$