When it is 154 m above the ground, a rocket traveling vertically upward at a con

Question

When it is 154 m above the ground, a rocket traveling vertically upward at a constant 7.80 m/s relative to the ground launches a secondary rocket at a speed of 13.8 m/s at an angle of 59.0 degrees above the horizontal, both quantities being measured by an astronaut sitting in the rocket. Air resistance is too small to worry about.

A.) Just as the secondary rocket is launched, what are the horizontal and vertical components of its velocity relative to the astronaut sitting in the rocket?

B.) Just as the secondary rocket is launched, what are the horizontal and vertical components of its velocity relative to Mission Control on the ground?

C.) Find the initial speed of the secondary rocket as measured by Mission Control.

D.) Find the launch angle of the secondary rocket as measured by Mission Control.

E.) What maximum height above the ground does the secondary rocket reach?

Explanation / Answer

Height of the rocket when it Launches the secondary rocket h = 145 m Relative velocity of the rocket with respect to earth (or) velocity of the rocket relative to earth v1 = 7.8 m/s Angel of projection of secondary rocket above the horizontal theta = 59 degree a) i)Velocity of the secondary rocket with respect to the astronaut sitting in the rocket v2 = 13.8 m/s horizontal component of velocity of the secondary rocket relative to the astronaut sitting v2x = v2 cos 59 = 13.8*cos59 = 7.107 m/s Vertical component of velocity of the secondary rocket relative to the astronaut sitting in the rocket v2y = v2 sin 59 = 13.8*sin59 =11.829 m/s ii)Velocity of the secondary rocket relative to the Mission control one the ground (or earth ) v3 = v1 +v2 = 7.8 + 13.8 = 21.6 m/s Horizontal component of velocity of the secondary rocket relative to the mission control on the ground v3x = v3 cos 59 = 21.6*cos59 = 11.125 m/s ……………. (1) Vertical component of velocity of the secondary rocket relative to the mission control on the ground v3y = v3 sin 59 = 21.6*sin59 = 18.515 m/s ………….. (2) b)The initial velocity of the secondary rocket as measure by mission control vbar = v3xi^ + v3yj^ (or) vbar = 11.125i^ + 18.515j^ Initial speed of he secondary rocket as measured by mission control mod(vbar) = sqrt(11.125^2+18.515^2) = 21.6 m/s Initial speed of the secondary rocket as measured by mission control is 21.6 m/s Since , the rocket travel vertically, Launch angle of the secondary rocket as measured by mission control = 59 degree c)Initial velocity of the secondary rocket relative to the earth v0 = 21.6 m/s Angle of projection theta = 59 degree In case of oblique projection , Maximum height H = vo^2sin^2(theta)/2g = 21.6*21.6*sin^2(59)/2*9.8 = 17.489 m Maximum height of the secondary rocket above the ground = [ height of the rocket when it fires secondary rocket ] + [H] = 145 + 17.489 = 162.489 m

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