an arrow is fired directly horizontal off a cliff that is 10.0 meters tall with a velocity of 65.5 m/san arrow is fired directly horizontal off a cliff that is 10.0 meters tall with a velocity of 65.5 m/s. How long is the arrow in the air?
10m = 0.5(9.8m/s^2)t^2
0.5*9.8=
h = o.5g*t^2
h = 10 m.
g = +9.8 m/s^2
Solve for t.
I don't know
Well, if the arrow is fired horizontally, it means it has no vertical velocity at the beginning. So, let's think about how long it will take for the arrow to fall from a height of 10.0 meters. We can use the good old equation for vertical displacement under constant acceleration, which is y = (1/2)gt^2, where y is the vertical displacement, g is the acceleration due to gravity (approximately 9.8 m/s^2), and t is the time.
In this case, the displacement of the arrow is -10.0 meters (negative because it's falling downwards), and we can solve for t. Plugging in the values, we get -10.0 = (1/2)(9.8)t^2. Solving this equation gives us t ≈ 1.43 seconds.
Now, since the arrow was fired horizontally, it will take the same amount of time to travel horizontally as it does to fall vertically. So, the arrow will be in the air for approximately 1.43 seconds.
Just to clarify, though, the arrow won't magically grow longer or change its length while in the air. It will maintain its original size. But I get what you mean!
To find out how long the arrow is in the air, we need to calculate the time it takes for the arrow to fall from the cliff to the ground.
First, we need to determine the time it takes for the arrow to reach the ground when fired horizontally. The horizontal velocity doesn't affect the time of flight in this case, as there is no acceleration horizontally.
The vertical motion of the arrow follows the equation:
y = y_0 + v_0y * t + (1/2) * g * t^2
Where:
y = final vertical position (0 meters, because the arrow reaches the ground)
y_0 = initial vertical position (10.0 meters)
v_0y = initial vertical velocity (0 m/s, as the arrow is fired horizontally)
g = acceleration due to gravity (-9.8 m/s^2, assuming no air resistance)
t = time of flight
Plugging in the values, we get:
0 = 10.0 + 0 * t + (1/2) * (-9.8) * t^2
Simplifying further:
0 = 10.0 - 4.9 * t^2
Now, we can solve for t by rearranging the equation:
4.9 * t^2 = 10.0
Dividing both sides by 4.9:
t^2 = 2.04
Taking the square root of both sides:
t = √(2.04)
t ≈ 1.43 seconds
Therefore, the arrow is in the air for approximately 1.43 seconds.