0
Am arrow is shot straight up at an initial velocity of 250 m s how long will it take to hit the ground?
Assuming no air resistance, the arrow will take approximately 5 seconds to hit the ground because it will reach its maximum height before falling back down due to gravity. The total time for the arrow to travel up and back down is twice the time it takes to reach the maximum height.
What else can I help you with?
An arrow is shot straight up at an initial velocity of 200 ms. How long will it be in the air before beginning to fall?
The time the arrow will be in the air before beginning to fall can be calculated using the formula t = (final velocity - initial velocity) / acceleration. Since the arrow is shot straight up, the final velocity at the top of its flight is 0. Given the initial velocity of 200 ms and acceleration due to gravity of -9.81 m/s^2, the time in the air before beginning to fall is approximately 20.4 seconds.
If an archer shoots an arrow straight up with an initial velocity magnitude of 100.0 ms After 5.00 seconds the velocity is 51.0 ms. At what rate is the arrow decelerated?
The average deceleration of the arrow can be calculated using the formula: average deceleration = (final velocity - initial velocity) / time. Plugging in the values gives an average deceleration of (51.0 - 100.0) / 5.00 = -9.8 m/s^2. This negative value indicates that the arrow is decelerating due to the acceleration of gravity.
A arrow is shot strait up at an initial velocity of 250meters per seconds How long will it take to hit the ground?
To calculate the time it takes for the arrow to hit the ground, we need to consider the vertical motion of the arrow. The time taken for an object to fall back to the ground can be determined using the kinematic equation: h = (1/2)gt^2, where h is the initial height, g is the acceleration due to gravity (approximately 9.81 m/s^2), and t is the time. In this case, the initial velocity is upwards, so the initial height will be 0. Using the equation, we can determine the time it takes for the arrow to hit the ground.
An arrow shot straight up into the air reached height of 75m with what velocity did it leave the bw how long was the arrow in the air?
To find the initial velocity of the arrow, you can use the equation Vf^2 = Vi^2 + 2gh, where Vf is the final velocity (0 m/s at the top of the flight), Vi is the initial velocity, g is the acceleration due to gravity, and h is the height reached (75m). Solve for Vi to get the initial velocity. To find the time the arrow was in the air, you can use the equation h = Vit - 0.5g*t^2, where t is the time in the air. Plug in the known values to solve for t.
An arrow is shot straight up at an initial velocity of 200 meters per seconds. How long will it be before beginning to fall?
The arrow will begin to fall when its velocity becomes negative, which will happen after it reaches its maximum height and starts to descend. The time it takes for the arrow to reach its peak height can be calculated using the formula: time = (final velocity - initial velocity) / acceleration. After reaching the peak, the arrow will take the same amount of time to fall back down.
An arrow is shot straight up at an initial velocity of 250ms How long will it take to hit the ground?
This is a velocity question so u need to use uvaxt
An arrow is shot straight up at an initial velocity of 200 ms. How long will it be in the air before beginning to fall?
The time the arrow will be in the air before beginning to fall can be calculated using the formula t = (final velocity - initial velocity) / acceleration. Since the arrow is shot straight up, the final velocity at the top of its flight is 0. Given the initial velocity of 200 ms and acceleration due to gravity of -9.81 m/s^2, the time in the air before beginning to fall is approximately 20.4 seconds.
An arrow in flight has an initial velocity of 65 meters per second and 10 seconds later it has a velocity of 35 meters per second Which is the acceleration of the arrow?
Acceleration of the arrow is -3m/s2A = (velocity minus initial velocity) / time
If an archer shoots an arrow straight up with an initial velocity magnitude of 100.0 ms After 5.00 seconds the velocity is 51.0 ms. At what rate is the arrow decelerated?
The average deceleration of the arrow can be calculated using the formula: average deceleration = (final velocity - initial velocity) / time. Plugging in the values gives an average deceleration of (51.0 - 100.0) / 5.00 = -9.8 m/s^2. This negative value indicates that the arrow is decelerating due to the acceleration of gravity.
A arrow is shot strait up at an initial velocity of 250meters per seconds How long will it take to hit the ground?
To calculate the time it takes for the arrow to hit the ground, we need to consider the vertical motion of the arrow. The time taken for an object to fall back to the ground can be determined using the kinematic equation: h = (1/2)gt^2, where h is the initial height, g is the acceleration due to gravity (approximately 9.81 m/s^2), and t is the time. In this case, the initial velocity is upwards, so the initial height will be 0. Using the equation, we can determine the time it takes for the arrow to hit the ground.
An arrow shot straight up into the air reached height of 75m with what velocity did it leave the bw how long was the arrow in the air?
To find the initial velocity of the arrow, you can use the equation Vf^2 = Vi^2 + 2gh, where Vf is the final velocity (0 m/s at the top of the flight), Vi is the initial velocity, g is the acceleration due to gravity, and h is the height reached (75m). Solve for Vi to get the initial velocity. To find the time the arrow was in the air, you can use the equation h = Vit - 0.5g*t^2, where t is the time in the air. Plug in the known values to solve for t.
An arrow is shot straight up at an initial velocity of 200 meters per seconds. How long will it be before beginning to fall?
The arrow will begin to fall when its velocity becomes negative, which will happen after it reaches its maximum height and starts to descend. The time it takes for the arrow to reach its peak height can be calculated using the formula: time = (final velocity - initial velocity) / acceleration. After reaching the peak, the arrow will take the same amount of time to fall back down.
An arrow is shot straight up at an initial velocity of 250 ms How long will it take the arrow to hit the ground?
Assuming no air resistance, the time it takes for the arrow to hit the ground is approximately 5 seconds. This is calculated using the kinematic equation: ( t = \frac{2v_0}{g} ), where ( v_0 = 250 , m/s ) and ( g = 9.81 , m/s^2 ) is the acceleration due to gravity.
When an archer shoots an arrow straight up with an initial velocity magnitude of 100.0 ms. After 5.00 s the velocity is 51.0 ms. At what rate is the arrow decelerated?
The change in velocity is 51-100 = -49 m/s This occurred over a period of 5 seconds so The (negative) acceleration - aka - deceleration is (-49 m/s)/(5 s) = -9.8 m/s²
Why an arrow shot horizontally and an arrow dropped straight down from the same height will hit the ground at the same time?
Both arrows will hit the ground at the same time because the force of gravity acts equally on both arrows, regardless of their initial horizontal or vertical motion. The vertical component of the horizontally shot arrow's motion does not affect the time it takes to fall to the ground.
When an arrow is used to show velocity what does the arrow tell you?
The arrow's direction indicates the velocity's direction, while the arrow's length represents the velocity's magnitude.
What happens when you shoot an arrow straight in the air?
An arrow shot vertically into the air will lose velocity and reverse direction. When it begins to descend, the fletching will quickly cause it to re-orient with the point downward. It will accelerate until it reaches its terminal velocity. Assuming an atmospheric density of 1.3 kg/m^3, and an arrow with a drag coefficient of 1.2, a weight of 0.23 N (200 grains), cross-sectional area of 23.48 mm^2, I calculate a terminal velocity of 112.6 m/s. That's about the same as the initial velocity when shot from a compound bow! One question remains, however: does the arrow have enough time during its descent to reach terminal velocity? That would depend on how high you shot it in the first place. Roughly speaking, though, the faster it was going when it left the bow, the faster it will be going when it reaches the ground.
