What was the initial speed of the archer's arrow?

[jwang023]

Homework Statement

An archer shoots an arrow horizontally at a target 16 meters away. The arrow is aimed directly at the center of the target, but it hits 57 centimeters lower.

What was the intitial speed of the arrow?

Homework Equations

The Attempt at a Solution

 

[rl.bhat]

Why the arrow hit the lower point?
Here horizontal velocity of the arrow remains constant. But the vertical velocity increases starting with zero.
Can you find the time t taken by arrow to fall 57 cm lower?
This is the time taken by the arrow to hit the target. Distance of the target is known. Time is known. Find the initial velocity.

 

[kerimek]

To solve this problem, we can use the equation for horizontal displacement, x = v*t, where x is the distance traveled, v is the initial velocity, and t is the time. We can also use the equation for vertical displacement, y = -1/2*g*t^2 + v0*t + y0, where g is the acceleration due to gravity, v0 is the initial velocity in the y-direction, and y0 is the initial height.

Since the arrow is shot horizontally, the initial velocity in the y-direction is 0, and the initial height is also 0. We can also assume that the time taken for the arrow to reach the target is the same as the time taken for the arrow to fall 57 centimeters. Therefore, we can set x = 16 meters and y = -0.57 meters in the equations.

Solving for v in the horizontal displacement equation, we get v = x/t. Plugging in the values, we get v = 16/t.

In the vertical displacement equation, we can substitute the value of v from the horizontal displacement equation. This gives us the equation -0.57 = -1/2*g*t^2 + (16/t)*t. Simplifying, we get -0.57 = -1/2*g*t^2 + 16.

Using the quadratic formula to solve for t, we get t = 0.256 seconds.

Substituting this value of t back into the horizontal displacement equation, we get v = 16/0.256 = 62.5 meters per second. Therefore, the initial speed of the arrow was 62.5 meters per second.