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.
When a softball is thrown, it has a positive acceleration because its velocity is increasing with time as it moves through the air.
To find the initial velocity with which the object is thrown upwards, we can use the kinematic equation for vertical motion: ( v = u + at ), where ( v ) is the final velocity (0 m/s at the highest point), ( u ) is the initial velocity, ( a ) is the acceleration due to gravity (-9.81 m/s²), and ( t ) is the time taken to reach the highest point (half the total time to reach the ground, so 2 seconds in this case). Rearranging the equation to solve for ( u ), we get ( u = v - at ). Plugging in the values, we get ( u = 0 - (-9.81 m/s² * 2 s) = 19.62 m/s ). Therefore, the object was thrown upwards with an initial velocity of 19.62 m/s.
Disturbed equilibrium refers to a state where a system that was previously in balance or stable has been disrupted or thrown off-balance. This can result in changes or fluctuations within the system until a new equilibrium is reached. Factors such as external influences or internal changes can lead to a disturbed equilibrium in various systems, such as in ecological, physical, or economic contexts.
In order for a body to escape the gravitational pull of the Earth, it needs to be thrown up with an initial velocity equal to or greater than the escape velocity of around 11.2 km/s. This velocity allows the object to overcome the gravitational pull of the Earth and continue traveling away from it indefinitely.
The impact velocity of a rock thrown horizontal from a cliff depends on two things, the initial speed of the rock (vi) and the height of the cliff (h). The final velocity (impact velocity) is represented by vfFor this formula, air resistance is neglected, and acceleration due to gravity is assumed to be 9.8 m/s2. The acceleration is positive here because down is being treated as the positive direction. You will get the same result if you use negative 9.8 m/s2 and make the height negative. sqr() means square root.vf = sqr(19.6h + vi2)For example if the rock was thrown off a 3 meter high cliff at 20 m/s, the impact velocity would be sqr(19.6 x 3 + 202), which would be sqr(58.8 + 202), which would be 21.42 m/s.The angle relative to the ground is the inverse tangent of sqr(19.6h)/viwhich in this case is tan-1( sqr(19.6 x 3)/20), which is tan-1(7.67/20) which is 21.0 degrees.
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.
The velocity of the ball at its highest point is 0 m/s. At the highest point of its trajectory, the ball's vertical velocity slows to 0 before changing direction and starting to fall back down due to the force of gravity.
At the highest point, the velocity of an object thrown vertically into the air is momentarily zero as it changes direction. This is the point where it transitions from going upward to downward.
9.8
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.
If a ball is thrown vertically upward with a velocity of 160 ft/s, then its height after t seconds is s = 160t - 16t^2. If a ball is thrown vertically upward with a velocity of 160 ft/s, then its height after t seconds is s = 160t − 16t^2.
The instantaneous speed of an object at its highest point when thrown straight up in the air is zero. At the highest point, the object has momentarily stopped moving upwards and is just about to start falling back down due to gravity.
The velocity of a ball thrown upward at W ft/sec will gradually decrease due to gravity until it reaches its highest point (at which the velocity will be 0 ft/sec). After reaching the peak, the ball will then start to fall back down, increasing in velocity until it reaches the ground.
Yes, it is possible for the initial velocity to be different from zero when the final velocity is zero. For example, an object could be thrown upwards and come to a stop at its highest point, where the final velocity would be zero.
The ball is affected by the force of the earth's gravity.
When the stone reaches its highest point, its vertical velocity is zero and its kinetic energy is at a minimum. At this point, all its initial kinetic energy has been converted into potential energy due to the force of gravity acting on it.