<h2><em>C. translational motion</em></h2><h2><em>HOPE IT HELPS !!!!!</em></h2>
Answer:
The acceleration is 14.28 km/h^2
Explanation:
Step one:
Given data
initial speed u= 0 km/h
final speed v= 140km/h
time t= 9.8 seconds
Required
The acceleration of the car
Step two:
From a= v-u/t
substitute
a= 140-0/9.8
a=140/9.8
a=14.28 km/h^2
Answer:
Explanation:
I will try to use newton’s second law and its’ concept of spring. The detailed solution is shown it the documents and I also used some mathematical concept which is highlighted.
Sum the forces in the y (upward) direction




Applying the kinematic equations of linear motion we have that the displacement as a function of the initial speed, acceleration and time is



Again through the kinematic equation of linear motion that describes velocity as the change of displacement in a given time, we have to



Therefore the horizontal distance between the target and the rocket should be 38.83m