The acceleration of the ball is about 9.8 m/s^2, which is the acceleration due to gravity.
In the first 2 seconds, the velocity of the ball would be given by v = at, where a is the acceleration. Given it traveled 2 meters in 2 seconds, we can use the equation s = (1/2)at^2 to find the acceleration which is 1 m/s^2. So, after 3 seconds, the ball will travel an additional 3 * 1 = 3 meters.
Assuming you release it from a position of rest, you must multiply the time by the acceleration. The acceleration due to gravity near Earth's surface is approximately 9.8 meters/second squared.
The acceleration of the ball is 9.8 m/s^2, which is due to gravity acting on it.
The acceleration of the sepaktakraw ball after 1 second will be approximately 9.81 m/s^2, assuming the ball is under the influence of gravity. This is the acceleration due to gravity acting downwards on the ball.
The distance a rubber ball falls in 10 seconds will depend on the height from which it is dropped and the acceleration due to gravity. On Earth, neglecting air resistance, the general equation to calculate the distance fallen is: distance = 0.5 * acceleration due to gravity * time^2.
In the first 2 seconds, the velocity of the ball would be given by v = at, where a is the acceleration. Given it traveled 2 meters in 2 seconds, we can use the equation s = (1/2)at^2 to find the acceleration which is 1 m/s^2. So, after 3 seconds, the ball will travel an additional 3 * 1 = 3 meters.
If the ball started off stationary and the acceleration is the same for the 2 seconds, then the acceleration of the ball is:Speed / time = acceleration9 (m/s) / 2 (s) = 4.5 (m/s/s)Hope this helps.
Assuming you release it from a position of rest, you must multiply the time by the acceleration. The acceleration due to gravity near Earth's surface is approximately 9.8 meters/second squared.
The acceleration of the ball is 9.8 m/s^2, which is due to gravity acting on it.
The acceleration of the sepaktakraw ball after 1 second will be approximately 9.81 m/s^2, assuming the ball is under the influence of gravity. This is the acceleration due to gravity acting downwards on the ball.
The distance a rubber ball falls in 10 seconds will depend on the height from which it is dropped and the acceleration due to gravity. On Earth, neglecting air resistance, the general equation to calculate the distance fallen is: distance = 0.5 * acceleration due to gravity * time^2.
A car increasing its speed from 0 to 60 mph in 10 seconds is an example of acceleration. A ball thrown into the air, moving faster and faster as gravity pulls it downward, is also experiencing acceleration.
The height of the ball after 3 seconds can be calculated using the formula for free fall: ( h = h_0 - \frac{1}{2} g t^2 ), where ( h_0 ) is the initial height (80 meters), ( g ) is the acceleration due to gravity (approximately 9.81 m/s²), and ( t ) is the time in seconds. After 3 seconds, the height is ( h = 80 - \frac{1}{2} \times 9.81 \times (3^2) ), which simplifies to ( h = 80 - 44.145 ). Therefore, the height of the ball after 3 seconds is approximately 35.855 meters.
Since the derivative of velocity is acceleration, the answer would be technically 'no'. Here is why: v = 0 v' = 0 = a Or in variable form... v(x) = x v(0) = 0 v'(0) = 0 = a You can "trick" the derivative into saying that v'(x) = 1 = a (since the derivative of x = 1) and then stating v'(0) = 1 = a... but that is not entirely correct. Acceleration is a change over time and is measured at more then one point (i.e. the acceleration of this body of matter is y from time 1 to 5) unless using derivatives to form the equation of the acceleration line/curve. If an object has a constant acceleration of 1, then the velocity is constantly increasing over that time. Using the equation discussed above and looking at acceleration over time, at 0 seconds, acceleration is 0 and so is velocity, but from 0-1 seconds acceleration is 1 and velocity is 1 as well. 0-2 seconds, acceleration is 1, but velocity would be 2 (at the end of 2 seconds).
The force needed would be equal to the mass of the ball multiplied by the acceleration, so 1 kg * 1 m/s^2 = 1 Newton.
To calculate acceleration between 6 and 9 seconds, you need to find the change in velocity during that time interval and then divide it by the time taken. The formula for acceleration is acceleration = (final velocity - initial velocity) / time. Plug in the velocities at 6 seconds and 9 seconds into the formula to get the acceleration.
To calculate the velocity of the ball just before it hits the ground, we can use the equation of motion: velocity = acceleration x time. The acceleration due to gravity is approximately 9.8 m/s^2. Given the time of 3.0 seconds, we can plug these values into the equation to find the velocity. Therefore, the velocity of the ball just before it hits the ground is 29.4 m/s.