Question

An arrow is shot parallel to horizontal from a height of 1.4 meters. It has a velocity of 49 m/s.

Answer 1

The time taken for the arrow to hit the ground is 0.53 s.

The horizontal distance traveled by the arrow is 25.97 m.

The given parameters:

• Initial horizontal velocity, $v_x$ = 49 m/s
• Initial vertical velocity, $v_y$ = 0

What is time of motion?

• This is the time taken for a projectile to land on the plane of projection.

The time of motion is calculated as follows;

$h = v_yt + \frac{1}{2} gt^2\\\\h = 0 + \frac{1}{2} gt^2\\\\h = \frac{1}{2} gt^2\\\\t = \sqrt{\frac{2h}{g} } \\\\t = \sqrt{\frac{2\times 1.4}{9.8} }\\\\t = 0.53 \ s$

The horizontal distance traveled by the arrow is calculated as follows;

$X = v_x t\\\\X = (49\ m/s) \times (0.53 \ s)\\\\X = 25.97 \ m$

Learn more about time of motion here: brainly.com/question/2364404