Question

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.

Answer 1

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.