Q1.a toy rocket is launched with an initial velocity of 11.0 m/s in the horizontal direction from the roof of a 40.0 m -tall building. the rocket's engine produces a horizontal acceleration of (1.60 m/s3)t, in the same direction as the initial velocity, but in the vertical direction the acceleration is g , downward. air resistance can be neglected. part a) what horizontal distance does the rocket travel before reaching the ground? write the answer with the unit.

The horizontal movement of the rocket is 11m/s, with an acceleration of 1.6m/s². The vertical movement will be downward, with an initial velocity of zero (it was shot horizontally) and a negative acceleration of g (-9.8m/s²)

To see how far the rocket traveled before hitting the ground, let's first figure out the time t at which the rocket hit the ground:
The formula for distance is d= vt + (1/2)at² ,
Where v=initial velocity, d=distance traveled, a=acceleration, and t=time
We want to find how long it took to travel 40 meters (height above the ground), given an initial velocity of 0 and negative acceleration of 9.8
Plugging into the equation:
40 = 0(t) + (1/2) (9.8) (t²)          Multiply both sides by (2/9.8)
8.16 = t²                                    Square root of both sides
t=  2.85

The rocket traveled for 2.85 seconds before hitting the ground. Plug this number into our distance formula to find horizontal distance
d= vt + (1/2)at² 
d = 11 (2.85) + (1/2) (1.6) (2.85²)
Remember that initial horizonal velocity is 11m/s and horizontal acceleration is 1.6m/s²
Simplify:
d= 31.35 + .8 * 8.16
d = 37.87

The object traveled 37.87 meters before hitting the ground. 

Ithink they do have the same properties just in less quantities i think