Question

An arrow is aimed horizontally, directly at the center of a target 20 m away. the arrow hits 0.050 m below the center of the target. neglecting air resistance, what was the initial speed of the arrow?

Answer 1

Answer:

200 m/s

Explanation:

h = 0.05 m, x = 20 m, uy = 0, g = 10 m/s^2

Let the speed in horizontal direction is u and the time taken is t.

h = uy t + 1/2 gt^2

0.05 = 1/2 x 10 x t^2

t = 0.1 s

Horizontal distance = horizontal speed x time

x = u x t

20 = u x 0.1

u = 200 m/s

Answer 2

The initial speed of the arrow is 200 m/s.

Given the following data:

• Horizontal distance = 20 meters

• Horizontal displacement, h = 0.050 meters.

• Vertical initial speed, Uy = 0 m/s

We know that acceleration due to gravity (a) is equal to 10 meter per seconds square.

To find the initial speed of the arrow:

First of all, we would use the second equation of motion to find the time taken during the horizontal displacement:

$h = U_y + \frac{1}{2}at^2\\\\0.050 = 0 + \frac{1}{2}(10)t^2\\\\0.050 = 5t^2\\\\t^2 = \frac{0.050}{5} \\\\t^2 = 0.01\\\\t = \sqrt{0.01}$

Time, t = 0.1 seconds.

Now, we would determine the initial speed of the arrow by using the formula;

$Initial \; speed = \frac{Horizontal \; distance}{Time}\\\\Initial \; speed = \frac{20}{0.1}$

Initial speed, Vo = 200 m/s

Therefore, the initial speed of the arrow is 200 m/s.

Read more: https://brainly.com/question/8898885