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

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

step-by-step explanation:

we are given

we have to simplify it

firstly, we will factor it

now, we can cancel like terms

now, we can combine it

so, we get

i don’t know what the answer is i wish i