Wiki User
∙ 13y agoIts acceleration is always the same - the acceleration of gravity at 32 ft/sec/sec - no matter what distance it is during drop, until it hits the ground.
Wiki User
∙ 13y agoAverage acceleration during a time interval = (change in speed) divided by (time for the change) =(25) / (5) = 5 meters per second2
Acceleration is the speed variation at a given time. Thus you must calculate the speed variation: Final speed - Initial speed = 25m/s-0m/s=25m/s The conscious then is performed over time.. t=5s Finally, the acceleration is a=(25m/s)/5s=5m/s2
Acceleration = (change in speed) divided by (time for the change) = (25) / (10) = 2.5 meters per second2
Assume the question should read, "If a triangle has an area of 100 sq feet and the base measures 25 feet then what is the height?"Area of a triangle = 1/2base x height.100 = 1/2 x 25 x height : height = 2 x 100 ÷ 25 = 8
Acceleration = (change in speed) divided by (time for the change)Acceleration = (25 - 0)/(3) = 81/3 miles per second2Don't try this at home. That acceleration is about 1,368 G's. You can not survive it!
The acceleration of the ball can be calculated using Newton's second law, which states that acceleration is equal to the force applied divided by the mass of the object. In this case, the acceleration would be 25 N divided by the mass of the ball in kg.
If we assume the time it takes for the ball to stop is directly proportional to the height it is dropped from, we can set up a proportion based on the given information. From the given data, we have the ratio of time to height as 11/1 = 25/2. Therefore, if we continue this ratio, the time it would take to stop if dropped from 3 feet would be 55 seconds.
Using Newton's second law (F=ma), the acceleration of the ball can be calculated by dividing the force (25 N) by the mass of the ball (0.3 kg). The acceleration of the ball would be 83.3 m/s^2.
To find the acceleration of the ball, we can use Newton's second law of motion: ( F = ma ). Given that the force is 25 N and the mass is 0.3 kg, we can rearrange the formula to find the acceleration: ( a = \frac{F}{m} = \frac{25}{0.3} \approx 83.3 , m/s^2 ). The acceleration of the ball is approximately 83.3 m/s^2.
The force needed to accelerate a 25 kg bowling ball would depend on the desired acceleration. Newton's second law states that force equals mass multiplied by acceleration (F = ma). If you specify the acceleration, the force required can be calculated using this formula.
To find the acceleration of the ball, you need to use Newton's Second Law, which states that acceleration is equal to the force applied divided by the mass of the object. In this case, the acceleration of the 0.30 kilogram ball that is hit with a force of 25 Newtons would be 83.3 m/s^2.
The force required to accelerate a 25 kg bowling ball can be calculated using the equation F = ma, where F is the force, m is the mass of the bowling ball, and a is the acceleration. If the acceleration is given, you can plug in the numbers to find the force needed.
The acceleration due to gravity is approximately 9.81 m/s^2 on Earth. When an object falls from a height of 25 m, it will experience this gravitational acceleration as it accelerates towards the ground.
At the point of impact, since force = mass x acceleration, acceleration = 25/0.5 = 50 N/kg If the catcher exerts a force of 25 newtons against the 0.5 kg ball, then he will cause it to accelerate at the rate of 50 m/sec-squared. If he happens to exert the force in the direction opposite to the velocity of the ball ... a common occurrence for a catcher ... then the acceleration is also opposite to the velocity of the ball, and the ball slows down.
Average acceleration during a time interval = (change in speed) divided by (time for the change) =(25) / (5) = 5 meters per second2
If we knew from what height the ball, when dropped, would reach its terminal velocity, and if we knew the percentage of rebound the ball would give, we could then be certain. I can only guess that a basketball will rebound approximately 75% of the height from which it is dropped, and if the height at which it would reach terminal velocity is maybe 300 feet, the ball would bounce back up to 225 feet. Just a guess! A basketball has an elasticity (or "bounciness") of about 56 percent.I'm not sure there's a theoretical limit. In practice, of course, there would be one: when the velocity of the ball impacting the ground is so great the ball explodes rather than bouncing. But you'd have to fire it out of some kind of basketball cannon to get it moving that fast.The official standard for ball inflation is that the ball should bounce roughly 75% of its drop height (specifically, between 49" and 54") when dropped from 6 feet. If you're referring to just the height a dropped ball could bounce and you're not throwing it down with some kind of basketball-downward-hurling machine, you could calculate the theoretical bounce height by figuring out what terminal velocity is for a basketball, calculating how high you'd have to drop it from (assuming no atmosphere) to achieve that velocity, and then multiplying by 0.75. I'm not going to do it for you, because I'm not actually all that interested in the answer, but that's how you could do it if you are.
1.6 ft