A soccer field is viewed from above, while a ball is kicked eastward with an initial speed of 10.0 m/s. The ball experiences a constant eastward acceleration of 0.75 m/s 2 over a distance of 20.0 m. Immediately after this 20.0 m distance, the ball is kicked northward in a manner that does not change its speed at that point, just its direction. The ball is allowed to come to a complete stop. How far north does the ball travel before stopping if the acceleration after this kick is 1.15 m/s 2

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nuhulawal20 nuhulawal20 · Answer 1 · 2021-02-19T02:22:24+01:00

Answer:

the ball travelled approximately 60 m towards north before stopping

Explanation:

Given the data in the question;

First course : [tex]a_{x}[/tex] = 0.75 m/s², [tex]d_{x}[/tex] = 20 m, [tex]u_{x}[/tex] = 10 m/s

now, form the third equation of motion;

v² = u² + 2as

we substitute

[tex]v_{x}[/tex]² = (10)² + (2 × 0.75 × 20)

[tex]v_{x}[/tex]² = 100 + 30

[tex]v_{x}[/tex]² = 130

[tex]v_{x}[/tex] = √130

[tex]v_{x}[/tex] = 11.4 m/s

for the Second Course:

[tex]u_{y}[/tex] = 11.4 m/s, [tex]a_{y}[/tex] = -1.15 m/s², [tex]v_{y}[/tex] = 0

Also, form the third equation of motion;

v² = u² + 2as

we substitute

0² = (11.4)² + (2 × (-1.15) × [tex]d_{y}[/tex] )

0 = 129.96 - 2.3[tex]d_{y}[/tex]

2.3[tex]d_{y}[/tex] = 129.96

[tex]d_{y}[/tex] = 129.96 / 2.3

[tex]d_{y}[/tex] = 56.5 m

so;

|d| = √( [tex]d_{x}[/tex]² + [tex]d_{y}[/tex]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