At what point does a baseball reach its highest speed when thrown. Please answer the 3 questions in the details?

At what point does a baseball reach its highest speed when thrown. Please answer the 3 questions in the details... Lets say that randy johnson can throw a consistent 105mph fastball. If he was to be in a field, and throw the ball as fast as he could, at what point (or distance) would the ball reach it's point of climax. I would guess that the ball is still increasing speed when it hits the catchers glove 60 feet away.

When does the ball climax and when does it slow down? How far will the 105mph fastball travel before it drops to the ground? Asked by dwdrums 52 months ago Similar questions: point baseball reach highest speed thrown answer questions details Sports & Recreation > Baseball.

Similar questions: point baseball reach highest speed thrown answer questions details.

Ball starts decelerating immediately I don't understand what makes you think the ball would be *increasing* in speed after the pitcher releases it. Quite the contrary, the moment the ball leaves the pitchers finger tips, only its own momentum carries it forward. It is slowed by air resistance.

There is nothing propelling it. Therefore, a ball's maximum velocity is achieived the instant the ball is released by the pitcher. As for how far will the ball travel before hitting the ground, this is a gh School physics question, in the realm of something called "Projectile Motion".

Solving it requires a couple understandings and a couple assumptions. First, we assume constant gravity. Second, we assume no wind resistance (which makes this not a "real" answer, but the best answer you're going to get without a LOT more complicated physics).

Going along with that assumption is that the pitcher's "spin" on the ball has no effect. It is the ball's spin that makes a ball "sink" or "hang", but that only happens because of the relative air pressures on the top and bottom of the ball. The two understandings you need are that: 1) There is nothing changing the speed of the ball in the horizontal direction.It's constant, at, if you like, 105mph.

2) The speed of the ball in the vertical direction changes in each moment by the effect of gravity. So let's set this up as a physics problem. First, our variables: t is the time of flight.

Ie, the number of seconds after the ball was released by the pitcher dh(t) is the horizontal distance the ball acheived after t seconds dv(t) is the vertical distance the ball achieved after t seconds vh is the horizontal velocity (a constant 105 mph, which is roughly 47 meters per second) vv is the intial vertical velocity. We assume the pitcher threw perfectly parellel to the ground, so the intial vertical velocity is 0. Vv(t) is the vertical velocity after t seconds Okay.

So, because there is nothing affecting the horizontal speed, the horizontal distance can be gotten from the definition of speed (or velocity). Velocity = distance/time. Solve for distance: distance = time * Velocity.

So dh(t) = t * vh What this means, somewhat obviously, is that the horizontal distance of the ball is dependent on how many seconds the ball travels through the air. So what we need to find is how many seconds before the ball falls to the ground from the position at which it started.To do this, we use the equation for motion for constant negative acceleration: The distance in the vertical direction is the velocity times the time, minus one half of the acelleration times the time squared. That is: dv = vv*t - (gt^2)/2 So the question is, how far is the ball going to fall.

According to various sites (ESPN.Com, sportsline, etc), Randy Johnson is 6'10", which is 2.0828 meters. For the sake of simplicity, we'll assume that when he releases the ball, his hand is just below the top of his head, so the ball has to fall exactly 2 meters. So we plug in our values: dv = -2m (distance to travel, downwards) vv = 0m/s (initial vertical velocity) g = 9.8m/s^2 (acceleration due to Earth's gravity) -2 = 0*t - (9.8*t^2)/2 -4 = -9.8*t^2 4/9.8 = t^2 0.408163265 = t^2 t = 0.6389s So now we know that the ball will fall to the ground from 2 meters in the air in 0.6389 seconds.So now we can find out how far in the horizontal direction the ball will travel in that time: dh(t) = t * vh dh(0.6389) = 0.6389 * 47 distance in the horizontal direction = 30.0283 meters = 98.5180446 feet So there you have it.

A ball thrown perfectly horizontally at 105 mph, from a height of 2 meters, assuming no wind resistance, will travel about 98½ feet before hitting the ground. Sources: memories of aided by: http://en.wikibooks.org/wiki/gh_school_physics/Projectile_motion .

A thrown baseball is moving its fastests when it leaves the pitcher's hand. The "point of climax" is assumed to be the highest speed of the ball, and not the farthest distance off the ground. The only force that I am aware of that could continue to accelerate the ball after leaving the pitcher's hand is gravity.

Gravity would add a downward force to the ball which would not increase its horizontal speed by any measurable amount. After the ball leaves the pitcher's hand, assuming it is thrown perfectly horizontally, it would immediately begin slowing down due to wind resistance. If a ball is thrown very fast, and it passes by you, you can actually hear this resistance.It sounds like a leaking tire: ssssssssssssssssssssss.

This is slowing the ball down. A pitcher's curve and breaking pitches are caused by the rotation of the ball converting the horizontal velocity into vertical or diagonal velocity due to the atmospheric drag of the laces. The ball is not still accelerating when it hits the catcher's mit.

How far will a 105MPH fast ball travel before it hits the ground depends on the height of release. Assuming a 5ft high release point, horizontal release, ignoring deflection due to lace drag (breaking balls), and no wind resistance, it can be estimated as follows: 105MPH = 154 feet per second (1 mile = 5280 feet, 1 hour = 3660 seconds, 105 * 5280 / 3660 = 154) Gravity = 32 feet per second per second. Distance = 5 = 16 * t^ 2.

5/16 = t^2. T = .559 seconds.154 * .559 = 86 feet Air resistence will shorten this slightly. I hope this helps.

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OK, here goes The baseball would be at its fastest at the instant it left his hand. At this point he can no longer add any energy to the ball and the effects of aerodynamic drag begin to slow the ball down. Mind you the speed drop from hand to mitt is fairly small.

A 105 mph fastball takes about 1/3 second from release point to crossing the plate. This is not much time for the air to slow the ball down. 60.5 ft at 105 mph is .3928 seconds but the Big Unit is probably releasing the ball at about the 53-54 ft.Mark.

The distance the ball will travel is determined (ignoring any wind) by: The initial velocity The angle to the ground of the initial direction vector Altitude above the ground at release The coefficient of drag for a MLB baseball A ball thrown at a 45 degree angle will travel the farthest. A ball thrown straight up will have the longest time of flight. Sources: Talent on loan from God .

Fastest when it leaves the thrower's hand. The ball will be going the fastest when it leaves the thrower's hand. All the forces (gravity, wind resistance and surface friction (including the catcher's mitt) are working against the initial ball speed.

Think of it this way. An object with mass needs a force for it to move. It also needs a counteracting force in order to slow down, stop, or move in the opposite direction.

The only forces (or a component) that are on the baseball after it leaves the thrower's hand are acting in the opposite direction to the original throw. Once it is hit by a bat, that places a greater force on the ball (in the opposite direction) and the ball travels in the reverse direction to the initial throw.

The answer is... Assuming that the pitcher is throwing in a straight line and not on an arc, the fastest speed of the ball should be at the point of release. After that point, the ball is being slowed by resistance from the air. The same principle is in affect for a gunshot, the speed at which a bullet leaves the barrel of a gun is called the muzzle velocity, a common gun rating.In addition to air resistance, the ball and bullet are being effected by gravity -- which is pulling the ball and the bullet both to the ground.

Although gravity may have a negligable effect on the ball's speed, because it it pulling the ball to the ground at and acceleration of 32 feet per second squared, the main factor effecting the speed is the air resistance. How far the ball goes depends on the speed of the ball. A faster released ball will travel farther than a slower released ball.

As soon as the ball is released, gravity is pulling it down to the ground in a few seconds, so a ball traveling faster can go farther in those few seconds until it hits the ground. Balls traveling in a arc have their speed effected by gravity because if you throw the ball up in the air away from the earth, gravity is pulling it down until the ball speed reaches 0 and it will accelerate as it drops to the earth. Sources: My days in school .

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