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 get 93, but my hw says its wrong
same question with different numbers cant get the answer don't think that's the right equation to use
To determine the speed at which the arrow was shot, we can use the principles of projectile motion. The key information we have is the distance the arrow traveled horizontally (62.0 m) and the angle it made with the ground (2.00 degrees).
First, we need to find the initial vertical velocity (Vy0) of the arrow. Since the arrow was shot parallel to the ground, the vertical component of the initial velocity is zero.
Next, we'll find the initial horizontal velocity (Vx0) of the arrow. We can use trigonometry to determine this value. The horizontal component of the initial velocity can be calculated using the formula:
Vx0 = V * cos(theta),
where Vx0 is the horizontal component of the initial velocity, V is the initial velocity of the arrow, and theta is the launch angle.
In our case, theta is 2.00 degrees. So, we have:
Vx0 = V * cos(2.00).
Now, to find the initial velocity (V), we can use the formula for the range of a projectile:
R = (V^2 * sin(2theta)) / g,
where R is the range of the projectile, V is the initial velocity, theta is the launch angle, and g is the acceleration due to gravity (approximately 9.8 m/s^2).
We know the range R is 62.0 m, and theta is 2.00 degrees. Plugging these values into the formula, we have:
62.0 = (V^2 * sin(4.00)) / 9.8.
Solving for V, we can rearrange the equation to:
V^2 = (62.0 * 9.8) / sin(4.00).
Finally, we can take the square root of both sides to find V:
V = sqrt((62.0 * 9.8) / sin(4.00)).
Using a calculator, we can evaluate this expression and find that the speed at which the arrow was shot is approximately 73.0 m/s (rounded to three significant figures).
Trig: horizontal distance=62
vertical distance=62*tan2
time to fall vertical distance...
h=1/2 g t^2
62*tan2=1/2 g t^2 solve for t.
speed= 62/t