#### Answer

v(10)= 60 ft/sec

#### Work Step by Step

The height of the rocket t seconds after liftoff is $3t^{2}$ ft.
Using kinematics, we know that the velocity of the rocket is $\frac{d}{dt}$($3t^{2}$ ), since the derivative of a position function with respect to time gives the velocity function.
Differentiating the position functions gives us v(t)=2*($3t^{2-1}$)= 6t.
v(t)=6t. So at t=10 sec, the rocket is climbing at 6(10)=60 ft/sec.