When it is 139 m above the ground, a rocket traveling vertically upward at a con

Question

When it is 139 m above the ground, a rocket traveling vertically upward at a constant 7.20 m/s relative to the ground launches a secondary rocket at a speed of 13.9 m/s at an angle of 54.0 above the horizontal, both quantities being measured by an observer sitting in the rocket. Air resistance should be ignored.

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

Find the initial speed of the secondary rocket as measured by Mission Control. Speed is the magnitude of the velocity vector of the secondary rocket with respect to the ground.? Find the launch angle of the secondary rocket as measured by Mission Control.?

What maximum height above the ground does the secondary rocket reach?

This is a standard projectile motion problem but all quantities used in the formulas should be with respect to the reference frame, which in this case is the ground. So the initial position of the secondary rocket with respect to the ground is the launch height. Max height occurs when the y component of the velocity with respect to the ground is zero. Acceleration on the secondary rocket is -g. Use the most convenient of the kinematic equations - the one that does not involve time.

Explanation / Answer

initial height ho = 139 m

speed of the primary rocket = 7.2 m/s

speed wrt to the primary rocket = 13.9 m/s

angle = 54 deg. with the horizontal

vertical component Vy = 13.9 Sin54 = 11.24 m/s

vert velocity wrt ground = 7.2 +11.24 = 18.44 m/s

horz. velocity wrt to ground = 13.9 Cos54 = 8.17 m/s

resultant speed wrt to ground

              = sqrt(18.44^2 +8.17^2) = 20.17 m/s

angle of launch wrt ground = arctan(18.44/8.17) = 66.1 deg.

maximu height reached from launch position

                                  h = Vy^2/2g = 18.44^2/2*9,8 = 17.35 m

height from ground = ho +h = 139.0 +17.35 = 156.35 m

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