You are watching an archery tournament when you start wondering how fast an arrow is shot from the bow. Remembering your physics, you ask one of the archers to shoot an arrow parallel to the ground. Unfortunately the archer stands on an elevated platform of unknown height. However, you find the arrow stuck in the ground 62.0 m away, making a 2.00 degree angle with the ground.

How fast was the arrow shooting?

I used
h=1/2 g t^2
62*tan2=1/2 g t^2

speed= 62/t

I got .66 for time
and I got a speed of 93
but my hw says its wrong

I would carry things to three sig digits, on time, and speed.

nope i still get the wrong answer

your assuming that the hypotenuse is a straight line when in reality its a negative expontial curve so this wont work

To find the speed at which the arrow was shot, you can use the concept of projectile motion. You correctly started with the equation for vertical motion:

h = (1/2)gt^2

where h is the height of the platform, g is the acceleration due to gravity, and t is the time of flight. However, since the question asks for the speed, you need to find the horizontal component of the velocity.

To do this, let's analyze the problem:

1. The arrow traveled a horizontal distance of 62.0 m.
2. The angle made with the ground is 2.00 degrees.

Now, let's break down the motion into horizontal and vertical components:

1. Horizontal motion:
- The horizontal distance traveled is 62.0 m.
- The horizontal component of the velocity (Vx) remains constant during the trajectory.

2. Vertical motion:
- The arrow was shot parallel to the ground, which means the initial vertical velocity (Vy) is 0.
- The arrow traveled a horizontal distance, so the time of flight (t) will be the same for both horizontal and vertical components.

From the given information, we can calculate the time of flight using the horizontal distance and the angle:

tan(2°) = Vy/Vx
tan(2°) = 0/Vx (since Vy = 0)
Vx = 62.0 m

Now, you can use the time of flight (t) and the horizontal distance (62.0 m) to find the speed:

Speed = Distance/Time
Speed = 62.0 m / t

However, we still need to find the value of t.

To find t, we can use the vertical motion equation. The vertical displacement (h) is given by h = (1/2)gt^2, where g is the acceleration due to gravity.

Plugging in the values:

h = (1/2)gt^2
h = (1/2)(9.8 m/s^2)t^2 (assuming g = 9.8 m/s^2, the acceleration due to gravity)
h = (4.9 m/s^2)t^2

You mentioned that the arrow stuck in the ground at a distance of 62.0 m away, making a 2.00° angle with the ground. The vertical displacement h is given by h = Xtan(θ), where X is the horizontal distance and θ is the angle.

h = Xtan(θ)
h = (62.0 m)tan(2°)

Now we can set the two equations for h equal to each other:

(4.9 m/s^2)t^2 = (62.0 m)tan(2°)

Now, you can solve this equation to find the value of t. Once you have t, substitute it back into the equation for speed to find the answer.

It's essential to perform the numerical calculation correctly to get an accurate answer. Always double-check your calculations and make sure to use the appropriate units.