Answer:
the ball travelled approximately 60 m towards north before stopping
Explanation:
Given the data in the question;
First course : 
= 0.75 m/s², 
= 20 m, 
= 10 m/s
now, form the third equation of motion;
v² = u² + 2as
we substitute
² = (10)² + (2 × 0.75 × 20)
² = 100 + 30
² = 130
= √130
= 11.4 m/s
for the Second Course:
= 11.4 m/s, 
= -1.15 m/s², 
= 0
Also, form the third equation of motion;
v² = u² + 2as
we substitute
0² = (11.4)² + (2 × (-1.15) × 
)
0 = 129.96 - 2.3
2.3 = 129.96
= 129.96 / 2.3
= 56.5 m
so;
|d| = √( ² + ² )
we substitute
|d| = √( (20)² + (56.5)² )
|d| = √( 400 + 3192.25 )
|d| = √( 3592.25 )
|d| = 59.9 m ≈ 60 m
Therefore, the ball travelled approximately 60 m towards north before stopping
Split the operation in two parts. Part A) constant acceleration 58.8m/s^2, Part B) free fall.
Part A)
Height reached, y = a*[t^2] / 2 = 58.8 m/s^2 * [7.00 s]^2 / 2 = 1440.6 m
Now you need the final speed to use it as initial speed of the next part.
Vf = Vo + at = 0 + 58.8m/s^2 * 7.00 s = 411.6 m/s
Part B) Free fall
Maximum height, y max ==> Vf = 0
Vf = Vo - gt ==> t = [Vo - Vf]/g = 411.6 m/s / 9.8 m/s^2 = 42 s
ymax = yo + Vo*t - g[t^2] / 2
ymax = 1440.6 m + 411.6m/s * 42 s - 9.8m/s^2 * [42s]^2 /2
ymax = 1440.6 m + 17287.2m - 8643.6m = 10084.2 m
Answer: ymax = 10084.2m
Trick question? In order to have kinetic energy, an object must be moving. Therefore, in this case, kinetic energy would be 0. If it were asking about potential energy, it would be a different story.
PE=mgh
180=0.5*10*h
180=5h
h=180/5=36 m
