In an archery practice an arrow was realesed from a bow with a velocity of 20m/s it strikes and slopes find at what rate the aarrow was aaccelarated
Ah, I initially thought you had too little information, but in fact this is theory (or should be). Depends, if you are neglecting dissipative forces like drag, this is quite easy.
So in projectile motion (the arrow is a projectile), we like to separate our axes into their respective components. This leaves us with x and y components (no z axis).
If we set the coordinate system with the y axis increasing up into the sky, and the x axis increasing away from the archer, this should give us two components of acceleration.
Well, not sure if you guys have done forces, but like I said we are going to neglect dissipative forces like air drag. This should leave only one exact force which acts on the arrow once in projectile motion.
This is the force of gravity. Considering gravity likes to act down toward the Earth (think, does our coordinate system describe a direction pointing down to the Earth), that should be simple. If you figure out where gravity acts, but are wondering about the other component, then step back for second. If gravity is hypothetically the only force acting, what could that possible mean for the other axis component?
To find the rate at which the arrow was accelerated, we need to know the time it took for the arrow to strike the target and the initial distance between the arrow and the target. Unfortunately, the given information does not include these values, so we cannot determine the rate of acceleration.
To find acceleration in general, we can use the formula:
acceleration = change in velocity / time
However, this formula assumes that the initial and final velocities are known, as well as the time it takes to reach the final velocity. In this case, since the time and distance of the arrow's flight are not provided, we cannot calculate the acceleration accurately.
If you have additional information such as the time or distance of flight, please provide it, and I can help you determine the acceleration.