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They might seem like they have the same flight patterns, but they don't. A 12-6 curveball thrown by an over-hander will go from 12 to 6, but if the same is thrown by a 3/4 or side-armer, the ball will react a little more like a slider.

Q: How can a ball thrown vertically have the same time flight as a ball thrown at an angle when they have the same maximum height?

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Ignoring air resistance, I get this formula:Maximum height of a vertically-launched object = 1.5 square of initial speed/GI could be wrong. In that case, the unused portion of my fee will be cheerfully refunded.

The speed decreases when an object is thrown vertically up because of the force of gravity acting against the object's upward motion. As the object goes higher, the force of gravity slows it down until the object reaches its maximum height, where its speed momentarily becomes zero before accelerating back downward.

A ball has the greatest potential energy at its highest point, such as when it is held at the top of a hill or at its maximum height when thrown vertically upwards. This is because the gravitational force acting on the ball is at its maximum, giving it the highest potential energy.

An object thrown vertically up wards from the ground returned back to the ground in 6s after it was thown up if it reached a height of 12m calculate?

if the bal is thrown by making 45 degree angles. with the ground..it will travel maximum distance...

apply the formula: v2 - u2 = 2as. Here v= 0; u = 19.6m/s; a = -g ,find s and that's max. heigth

The body will continue to rise until the force of gravity acting against its motion brings it to a stop before it falls back down to its starting position due to gravity pulling it back down. The total time of flight and maximum height reached depend on the initial velocity of the body and the acceleration due to gravity.

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 you ignore air resistance, then they will reach their maximum height at the same time. In order not to ignore air resistance, you would need to know their shapes.

To estimate the maximum height a ball could be thrown, you would need to consider factors like the initial velocity of the throw, the angle at which the ball is released, and the force of gravity. Using kinematic equations, you can calculate the height based on these factors. Keep in mind that air resistance and other external factors may affect the actual height achieved.

A ball is thrown vertically upward with an initial speed of 20m/s. Two second later, a stone is thrown vertically (from the same initial height as the ball) with an initial speed of 24m/s. At what height above the release point will the ball and stone pass each other?