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
Initially, the velocity vector is
. At the same height, the x-value of the vector will be the same, and the y-value will be opposite (assuming no air resistance). Assuming perfect reflection off the ground, the velocity vector is the same. After 0.2 seconds at 9.8 seconds, the y-value has decreased by
, so the velocity is
.
Converting back to direction and magnitude, we get 
Answer:
They are 7.4m apart.
Explanation:
Here we have a parabolic motion problem. we need the time taken to land so:

considerating only the movement on Y axis:

Because we have a contant velocity motion on X axis:

and

the distance between them is given by:

Answer:
70.07 Hz
Explanation:
Since the sound is moving away from the observer then
and
when moving towards observer
With
of 76 then taking speed in air as 343 m/s we have


Similarly, with
of 65 we have

Now

v_s=27.76 m/s
Substituting the above into any of the first two equations then we obtain

Answer:
1. Huge wildfires
2. Deforestation
3. Reduced amount of aforestation, etc