0
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 the
value of the acceleration at any particular instant during the 10 seconds.
-- We're working entirely with scalars ... speed, not velocity, and magnitude of
acceleration ... since we don't know anything about the arrow's direction at any
time during the whole event.
What else can I help you with?
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 a cricket jumps off the ground with an initial vertical velocity of 4 ft per second afer how many seconds does the cricket land on ground?
To determine how long it takes for the cricket to land back on the ground after jumping with an initial vertical velocity of 4 ft per second, we can use the formula for the time of flight in projectile motion. The time to reach the maximum height is given by ( t = \frac{v}{g} ), where ( v ) is the initial velocity and ( g ) is the acceleration due to gravity (approximately 32 ft/s²). In this case, it takes ( t = \frac{4}{32} = 0.125 ) seconds to reach the peak. Since the time to ascend and descend is equal, the total time until the cricket lands back on the ground is ( 2 \times 0.125 = 0.25 ) seconds.
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
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
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.
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?
If you're willing to ignore the effect of air resistance, then the answer is as follows: The object's horizontal velocity remains constant (at least until it eventually hits the ground). The vertical component of the object's initial velocity ... call it V(i) ... is the (total initial velocity) multipled by the (sine of the initial angle above the horizontal). Beginning at the time of the toss, the magnitude of the vertical component of velocity is V = V(i) - 1/2gT2. T = number of seconds after the toss g = acceleration of gravity = approx 32 ft/sec2 or 9.8 m/sec2
If a basketball player throws a ball from 2.0m in the air towards the hoop 10m away If the hoop is 3.05m off the ground and the initial angle is 40 degrees what is the initial velocity?
The initial velocity of the ball can be calculated using the kinematic equation for projectile motion. By using the vertical component of velocity (V0y) and the time of flight, we can determine the initial velocity needed for the ball to reach the hoop. The velocity components are V0x = V0 * cos(θ) and V0y = V0 * sin(θ), where θ is the initial angle. The time of flight in this case is determined by the vertical motion of the ball, and it can be found by using the equation of motion for the vertical direction, considering the initial vertical velocity, the gravitational acceleration, and the vertical displacement of the ball. Once these values are calculated, the initial velocity can be computed by combining the horizontal and vertical components of the motion.
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?
*Assuming no air resistance*Horizontal component:velocity = distance / timeVx = 50 m / 3 sec = 16.67 m/sVertical component:We know at the top point in the trajectory, the vertical velocity = 0. This happens at half the time. We will use acceleration = -9.8 m/s^2Final velocity = Initial velocity + Acceleration x time0 m/s = Vy + -9.8 m/s^2 x 1.5 sVy = 14.7 m/sInitial velocity can be found using Pythagorean theorem and the angle of the kick can be found using trigonometry.V = 22.2 m/sAngle = 41.4 degrees.
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.
