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∙ 15y agoThe position of an object in a gravity field is given by: S = 1/2 * a *t ^2 + v * t + x So making the assumption that both are subject to Earth's gravity, a = - 9.81 m/s^2 (negative acting down). Assuming the top stone starts at rest, V_top = 0 m/s, V_bottom = 25 m/s. Taking the problem statement, x_top = 100 m and x_bottom = 0 m. If you want the time when they impact one another, set the two position equations equal and solve for t: 1/2* - 9.81 * t ^2 + 0 + 100 = 1/2 * -9.81 * t^2 + 25 * t + 0 So the acceleration terms cancel, and t = 4 seconds. To check, solve both position equations at t = 4 s. S = 1/2 * -9.81 * 4 ^2 + 0 + 100 = 21.52 m S = 1/2 * -9.81 * 4 ^ 2 + 25 * 4 + 0 = 21.52 m So there you go. The time is 4 seconds, the position is 21.52 m above the ground.
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∙ 15y agoAssuming that your units of velocity are in units/second Acceleration = (velocity 2 - velocity 1) / time Acceleration = (4.9 - 0) / 3 Acceleration =1.63 *With correct significant figures the answer is 2
The initial velocity is zero. In most basic physics problems like this one the initial velocity will be zero as a rule of thumb: the initial velocity is always zero, unless otherwise stated, or this is what you are solving for Cases where the initial velocity is not zero examples a cannon ball is shot out of a cannon at 50 mph a ball is thrown from at a speed of 15 mph etc
If the ball was dropped from a roof and hit the ground 3.03 seconds later, then when it hit the groundits velocity was 29.694 meters (97.42 feet) per second (rounded) downward.
In two seconds of fall, the speed increases 19.6 meters (64.4 feet) per second. The magnitude of velocity increases by that amount, while the direction of velocity doesn't change.
Acceleration = (change in velocity) / (time for the change)9.8 = (change in velocity) / (2 seconds)9.8 x 2 = change in velocity = 19.6 meters per second .Hint: The mass of the object and the height of the building are there just tothrow you off balance. You don't need either of them to answer the question.
No, both balls will hit the ground at the same time, assuming they are dropped from the same height and in a vacuum. The horizontal velocity does not affect the time it takes for an object to fall vertically due to gravity.
The minimum velocity required to project a body from a height h so that it will not fall back to Earth is called the escape velocity. This velocity must be equal to or greater than the escape velocity, which is derived using the principle of conservation of energy and the gravitational force equation. The escape velocity is given by the equation v = sqrt(2 * g * h), where v is the escape velocity, g is the acceleration due to gravity, and h is the height from which the object is projected.
I say NO. If you mean it is dropped and falls vertically. Discover Channel's "Myth Busters" tried to determine if a bullet would kill you if it was fired directly vertical and falls on its own. The bullet or penny would fall at terminal velocity which is about 120mph. However, they will tumble which slows them down more. This velocity and their mass is not enough to kill you.
No, both balls will hit the ground at the same time if they are dropped from rest or launched horizontally from the same height. The initial horizontal velocity does not affect the time it takes for an object to fall vertically to the ground.
When a ball is dropped, it starts with an initial velocity of zero. However, as it falls towards the ground, it accelerates due to gravity, causing its velocity to increase. Therefore, the velocity of the ball is non-zero as it falls towards the ground.
The height from which an object is dropped does not affect its average velocity. Average velocity depends on the overall displacement and time taken to achieve that displacement, regardless of the initial height of the object.
a missle is a kind of bomb# An object or weapon that is fired, thrown, dropped, or otherwise projected at a target; a projectile. # A guided missile. # A ballistic missile. # An object or weapon that is fired, thrown, dropped, or otherwise projected at a target; a projectile. # A guided missile. # A ballistic missile.
The bomb will fall vertically downward relative to the plane as both the bomb and the plane are in motion. The horizontal distance between the bomb and the plane will increase as the bomb falls due to the plane's forward velocity.
The bullet fired horizontally will hit the ground first, given that it has an initial horizontal velocity that keeps it moving forward from the moment it leaves the gun. On the other hand, the bullet dropped from the end of the barrel only has the force of gravity acting on it, causing it to fall vertically, which is slower than the horizontal motion of the fired bullet.
The final velocity of a dropped object can be calculated using the equation v = gt, where v is the final velocity, g is the acceleration due to gravity (approximately 9.81 m/s^2), and t is the time the object has fallen. Plugging in the values, the final velocity of a dropped object after falling for 3.0 seconds would be 29.43 m/s.
The velocity-time graph for a body dropped from a certain height would show an initial spike in velocity as the object accelerates due to gravity, reaching a maximum velocity when air resistance equals the force of gravity. After this, the velocity would remain constant, representing free fall with a terminal velocity. When the object hits the ground, the velocity suddenly drops to zero.
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