Archer shoots his arrow towards a target at a distance of 90 m and hits bullseye . Calculate the acceleration and time taken by the arrow to hit the target with a velocity of 90.0 m/s

acceleration = 45 m/s²

Time = 20 seconds

Explanation:

We are given;

Distance; s = 90m

Final velocity;v = 90 m/s

Since shot from rest, initial velocity;u = 0 m/s

So, to find the acceleration, we can use Newton's 2nd law of motion which is;

v² = u² + 2as

Where a is acceleration

Plugging in the values;

90² = 0² + 2a(90)

8100 = 180a

a = 8100/180

a = 45 m/s²

Now, for the time, we will use Newton's 1st law of motion which is;

v = u + at

Where t is time;

So;

90 = 0 + 4.5t

90 = 4.5t

t = 90/4.5

t = 20 seconds

see the picture for the vector

the magnitude of the vector by pythagorean theorem

r = √( (-10)^2 + (3)^2 ) = √109

magnitude = √109 units

the direction angle between vector and x-axis is

tan^(-1)(3/10) = 16.699

so the direction is approximately 16.7 degrees north of west