An archer shoots and arrow horizontally at a target 15m away. The arrow is aimed directly at the center of the target, but it hits 52 cm lower. What was the initial speed of the arrow?
I know it has something to do with a zero launch angle but none of the equations i use come out to the correct answer in the back of the text.
How far did it fall? ANS .52 cm
How long did it take to fall? ANS d=1/2 g t^2 solve for t.
What was the speed of arrow? 15/t
I got 4.607m/s but im not so sure.sorry
To find the initial speed of the arrow, we can use the horizontal distance it traveled and the vertical displacement it experienced.
First, let's find the time it took for the arrow to fall by using the equation for vertical displacement:
d = (1/2) * g * t^2
Here, d represents the vertical displacement or the distance the arrow fell, which is given as 52 cm or 0.52 m. g is the acceleration due to gravity, which is approximately 9.8 m/s^2. We need to solve this equation for t:
0.52 = (1/2) * 9.8 * t^2
Rearranging the equation:
t^2 = (2 * 0.52) / 9.8
t^2 = 0.106122
Taking the square root of both sides:
t = √(0.106122)
t ≈ 0.33 seconds
Next, let's find the horizontal speed or initial speed of the arrow. We know that the horizontal distance the arrow traveled is 15 m, and the time it took to hit the target is approximately 0.33 seconds. We can use the equation:
speed = distance / time
speed = 15 / 0.33
speed ≈ 45.45 m/s
Therefore, the initial speed of the arrow is approximately 45.45 m/s.