Yes, a ball is in equilibrium at the top of it's throw because there is a moment of no change, or equilibrium, when it is suspended in air.
No, to be in equilibrium all of the forces acting on an object have to cancel each other out. Which in this example, gravity is the only force.
no, it still has the force of gravity on it
The force of gravity.
Answer:Yes, but only instantaneously.Consider a thrown ball moving directly upward. At the highest point of its trajectory, the instanataneous velocity (the velocity at that precise instant) is zero even while the acceleration due to gravity remains non zero.
A rock has the same constant acceleration from the moment it leaves your hand until the moment it hits the ground. It doesn't matter whether you dropped it or threw it, or in what direction it left you. The acceleration is 9.8 meters (32.2 feet) per second2 directed downwards. That's the acceleration of gravity on earth. As you asked, let's say you tossed it straight upwards. A tiny instant before it reaches the exact top, it has a small upward speed. A tiny instant after it passes the exact top, it has a small downward speed. During that tiny space of time, its upward speed decreases and its downward speed increases. That's a downward acceleration in anybody's book.
What distance does it move vertically? Throw a ball straight up. Gravity will slow it down at the rate of 9.8 m/s each second. Example A ball is thrown up at a velocity of 49 m/s. Due to gravity it will slow down at the rate of 9.8 m/s each second. After 5 seconds the ball will stop, it is at its highest point; the velocity is 0 m/s. The initial velocity was 49 m/s; the final velocity is 0m/s.The average velocity is 24.5 m/s. Distance = Average velocity * time Distance = 24.5 m/s * 5 seconds = 122.5 m Here are the physics equations I used!! Acceleration due to gravity = g = -9.8m/s each second = -9.8m/s^2 (negative because gravitational force pulls down toward the center of the Earth!! Vfinal = Vinitial + (acceleration * time) 0 m/s = 49m/s+ (-9.8m/s^2 * time) -9.8 t = - 49m/s t = 5 seconds Average velocity = (49m/s + 0) ÷ 2 Average velocity = 24.5 m/s Distance = Average Velocity * time. Distance = 24.5 m/s * 5 seconds Distance = 122.5 m
Using the equation: x=vot+(1/2)at2 x: -39.2m (though the object is thrown upwards, its total displacement is just the amount it fell from the tower). vo: initial velocity t: 4 s a: -9.8m/s2 (assuming we're on the earth's surface) -39.2=vo(4 s)+(1/2) (-9.8m/s2)(4 s)2 ((78.4 m - 39.2 m)/ 4 s) = v0 So the initial velocity is 9.8 m/s.
A the highest point its velocity will be zero.
The acceleration is the acceleration of gravity, downwards, or 9.8m/s/s (32 ft/s/s). When ball is thrown straight up it has an initial velocity that is decreasing because of gravity; at the highest point velocity is zero but acceleration is always constant at gravity rate.
If it is thrown at an angle, at the top of its path, its vertical velocity will be zero, however its horizontal velocity will be the same as its initial horizontal velocity minus whatever loss in speed as a result of air friction at that point. We won't know what that is without more information.
9.8
0 zero
The highest point is the point where the ball's velocity transitions from upward to downward. At that instant, the ball's speed, velocity, momentum, and kinetic energy are all exactly zero.
When the stone reaches its highest point, earth's gravity ensures it has to come down.
At the highest point, the kinetic energy is least.
A projectile thrown with a greater velocity would travel a greater distance. Velocity is not just speed but direction as well.
A projectile thrown with a greater velocity would travel a greater distance. Velocity is not just speed but direction as well.
A projectile thrown with a greater velocity would travel a greater distance. Velocity is not just speed but direction as well.
For example, an object thrown upwards, when it is at its highest point. This situation is only possible for an instant - if the acceleration is non-zero, the velocity changes, and can therefore not remain at zero.