Calculate The Maximum Height To Which M1 Rises After The Collision.

What is the formula to calculate height in physics?

Generally height is measured not calculated. It can be calculated in certain conditions if you have the data necessary. For example, if you toss an object vertically into the air of mass, m, and velocity v, it will have an initial kinetic energy of 1/2 mv^2 and as it rises its kinetic energy will decline and its potential energy will increase. When it comes to a stop as some height, h, all its energy will be potential and none will be kinetic because it is stopped. The potential energy of an object of mass m at height h in a gravitational field g is mgh.Thus 1/2 mv^2 = mgh and we solve for h. m cancels from both sides then divide through by g and you get v^2/2g = h. Let’s check the units. v has units of meters/sec, so v^2 has units of meters squared/seconds squared. g has units of meters/second squared. So meters squared/seconds squared/meters/second squared and the seconds squared cancel out and meters squared divided by meters is meters. So the unites check and we have a formula which will predict the height the object will reach if it is initially going upwards at velocity v.This is just an example. In other situations the formula would be different.

A ball of mass 8 kg is dropped from a height of 10 m. What is the velocity with which it strikes the ground?

As an object falls from rest, its gravitational potential energy is converted to kinetic energy. Consider a mass m which is falling vertically under the influence of gravity.Potential Energy = Mass x Gravity x HeightPE = mghKinetic Energy = ½ mv2KE = ½ mv2By conservation of energyEnergy before = Energy afterInitial KE + Initial PE = Final KE + Final PEObject is falling from rest, therefore initial kinetic energy is zero. Once the object hit on the ground, height is zero, therefore no potential energy at ground level.We can rewrite above equationInitial PE = Final KEThe mass m on both sides tells us that the final velocity doesn’t depend upon the mass.Velocity just before the impact is

A body is thrown vertically upward and rises to a height, and comes back to the initial position. What is the total distance travelled by the body? What is the displacement of the body?

Since it was thrown up vertically up, and as it travels vertically up tiil it came to rest, distance travelled up is the height h.As it returned vertically down , through a distance h, the total distance is h + h = 2h.Since it was thrown up vertically up, and as it travels vertically up tiil it came to rest, displacement was the height h.As it returned vertically down , through a displacement -h , the total displacement is h -h = 0Or as the starting and finishing points are the same displacement is zero.

A ball falls from a height such that it strikes the floor of a lift at 10 m/s. If the lift is moving in the upward direction with a velocity 1m/s, then what is the velocity with which the ball rebounds after elastic collision?

The lift is moving in the upward direction with a velocity of 1m/s.The ball moves in the downward direction and hits the floor with a speed of 10m/s.Meaning the relative velocity of between ball and lift is 10m/s.Meaning the ball is travelling towards the lift floor at 9m/s (just before the collision) and lift floor is travelling towards the ball at 1m/s.Considering it a pure elastic collision, no energy is lost during the collision (which would be almost instantaneous). And the lift is too massive (compared to the ball) to get a change in its speed (assuming ideal condition), the speed of the ball just before and just after the collision would be same implying that the speed of the ball (just) after the collision would be 9m/s (upwards).

A 50 g ball is dropped from a height of 1.5 m. After striking the floor, the ball bounced back vertically upward, losing 20% of the initial energy. What is the velocity of the object during the first rebound?

Firstly, we need to use a SUVAT equation to calculate the velocity of the ball at the moment it impacts the floor, I’ll be using v^2 = u^2 + 2as, where ‘v’ is the final velocity, ‘u’ is the initial velocity, ‘a’ is the acceleration of the object, and ‘s’ is the distance travelled by the object. In this case, ‘v’ is the variable to be found, ‘u’ equals 0, ‘a’ equals 9.81, and ‘s’ equals 1.5.v^2 = 0 + 2(9.81)(1.5)v^2 = 29.43v = 5.42 metres per second (3 significant figures)Next, we need to calculate the kinetic energy gained during the fall and it’s maximum value at the moment of impact, to do so I’ll use the equation KE = 0.5 * mass * velocity^2. Where the mass equals 50 grams, which is 0.05 kilograms, and the velocity is 5.42 metres per second). This gives us:KE = (0.5)(0.05)(5.42 * 5.42)KE = 0.73441 JoulesAs the question states, there is a 20% energy loss during the first rebound.100% - 80% = 20%20% = 0.20.73441 * 0.8 = 0.587528 JoulesWe put this back into the Kinetic Energy equation, giving us:0.587528 = 0.5 * 0.05 * v^20.587528 / (0.5 * 0.05) = v^223.14112 = v^2v = -4.81 metres per second (3 significant figures)The answer is negative as we took the positive values of acceleration and velocity to be the direction of down towards the floorNote that this is at the instant the ball begins it’s vertical ascent after the first bounce and will also be the speed (not velocity) at the instant of impact on the second bounce.