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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?
Wiki User
∙ 9y ago
Acceleration of the arrow is -3m/s2
A = (velocity minus initial velocity) / time
Wiki User
∙ 9y ago
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If 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?
Magnitude of average acceleration = (change of speed) divided by (time for the change)Average 'A' = (35 - 65) / 10 = -30/10 = -3.5 meters per second2-- That's the average over the 10 seconds. We don't know anything about thevalue of the acceleration at any particular instant during the 10 seconds.-- We're working entirely with scalars ... speed, not velocity, and magnitude ofacceleration ... since we don't know anything about the arrow's direction at anytime during the whole event.
A rocket gose from 0 m per second to 560 m per second in the first 7 seconds of flight What is your acceleration?
a = m/s/s a = 560/1/7 a = 80m/s/s
How can vectors be applied to real life situation?
You use it when throwing an object at a target. Over any but a trivially short distance, gravity will pull the object downwards. So you aim higher than the target. To hit the target, the vector sum of the initial velocity and the downward acceleration experienced during the flight must be a vector aimed directly at the target.
A boy throws a rock straight up in the air It reaches the highest point of its flight after 2.5 seconds how fast was the rock going when it left the boys hand?
There is no way to tell unless a height is specified. Once you have that, you would divide the distance (height) by the time (2.5). Suppose 50 feet, and it took 5 seconds to reach that height. You would have 50/5 = 10 feet per second. given the acceleration of gravity is 9.81m/s2 and y=at2 then ymax = 9.81(2.5)2 OR 61.3125m At its highest point it has a velocity of zero. if 0=v0-at and a=9.81 and t=2.5 then v0 = 9.81(2.5) = 24.525 m/s
A stone is projected with a velocity of 58.8 ms at an angle of 30 degree from horizontal find the Time of flight maximum height and horizontal range?
A stone is thrown with an angle of 530 to the horizontal with an initial velocity of 20 m/s, assume g=10 m/s2. Calculate: a) The time it will stay in the air? b) How far will the stone travel before it hits the ground (the range)? c) What will be the maximum height the stone will reach?
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.
A ball thrown vertically upward returns to the starting point in 8 seconds.find its initial velocity.?
The total time of flight for a ball thrown vertically upwards and returning to its starting point is twice the time taken to reach maximum height. Therefore, the time taken to reach maximum height is 4 seconds. Given that the acceleration due to gravity is -9.8 m/s^2, using the kinematic equation v = u + at, where v is the final velocity (0 m/s at maximum height), u is the initial velocity, a is the acceleration due to gravity, and t is the time, you can solve for the initial velocity. Substituting the values, u = 9.8 * 4 = 39.2 m/s. Therefore, the initial velocity of the ball thrown vertically upward is 39.2 m/s.
If 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?
Magnitude of average acceleration = (change of speed) divided by (time for the change)Average 'A' = (35 - 65) / 10 = -30/10 = -3.5 meters per second2-- That's the average over the 10 seconds. We don't know anything about thevalue of the acceleration at any particular instant during the 10 seconds.-- We're working entirely with scalars ... speed, not velocity, and magnitude ofacceleration ... since we don't know anything about the arrow's direction at anytime during the whole event.
What quantities are zero throughout the flight of a projectile?
The vertical velocity at the highest point of the trajectory, the vertical displacement when the projectile returns to its initial height, and the vertical acceleration at the highest point are all zero throughout the flight of a projectile.
How do you find time of flight from initial velocity and an angle at which an object is thrown?
You can find the time of flight by using the formula: time of flight = (2 * initial velocity * sin(angle)) / gravitational acceleration. Input the initial velocity and angle at which the object is thrown into the formula to calculate the time it takes for the object to reach the same height as it was initially launched.
If a football player kicks a football so that it spends 3 seconds in the air and travels 50 m then find the velocity and angle of the kick?
Given that the time of flight is 3 seconds and the displacement is 50 m, you can use the kinematic equations to find the initial vertical and horizontal velocities. Then, you can calculate the magnitude of the velocity and the angle by using trigonometry.
A player kicks a soccer ball from ground level and sends it flying at an angle of 30 degrees at a speed of 26 ms. How long was the ball in the air Round the answer to the nearest tenth of a second.?
To calculate the time the ball was in the air, you can use the kinematic equation for projectile motion. The time of flight is given by the formula: time = 2 * initial velocity * sin(angle) / acceleration due to gravity. Plugging in the values (initial velocity = 26 m/s, angle = 30 degrees, acceleration due to gravity = 9.81 m/s^2), you can calculate the time to be approximately 2.4 seconds.
What is the relationship between vertical displacement and theoretical time of flight?
The vertical displacement of a projectile is directly related to the theoretical time of flight. The higher the vertical displacement, the longer the projectile will stay in the air before landing. This is because the time of flight is influenced by the initial vertical velocity and acceleration due to gravity acting on the projectile.
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.
What do you need to know to be able to determine how far a projectile travels horizontally?
To determine how far a projectile travels horizontally, you need to know the initial velocity of the projectile, the angle at which it was launched, and the acceleration due to gravity. Using these values, you can calculate the time of flight and then multiply it by the horizontal component of the initial velocity to find the horizontal distance traveled.
What do you need to determine the time that a projectile is in motion?
To determine the time a projectile is in motion, you need to know the initial velocity of the projectile, the angle at which it is launched, and the acceleration due to gravity. Using these parameters, you can calculate the time of flight using projectile motion equations.
You shoot an arrow into the air. Two seconds later the arrow has gone straight upward to a height of 33.0 m above its launch point. What is the arrow's initial velocity?
The arrow's initial velocity can be calculated using the equation: ( v_f = v_i + at ), where ( v_f = 0 ) (final velocity), ( v_i ) is the initial velocity, ( a = -9.8 , \text{m/s}^2 ) (acceleration due to gravity), and ( t = 2 , \text{s} ) (time taken to reach the maximum height). Solving for ( v_i ) gives: ( 0 = v_i - 9.8 \times 2 ), so ( v_i = 19.6 , \text{m/s} ).
