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∙ 15y agoThe question seems ambiguous. Are you saying that both objects reach the water exactly one second after the first object is dropped? Are you saying that the second object was thrown exactly one second after the first object was dropped? I'm assuming the idea is that the objects reach the water at the same moment, and that you are ignoring the effect of air friction on the objects. A2 I assume object 2 is thrown 1 sec after object 1 is dropped and they both reach the bottom at the same time. Taking down as positive, the equation for each object is: L = (1/2)g(T - t1)^2 L= V(T-t2) + (1/2)g(T-t2)^2 Where T is the time they reach bottom. t1 & t2 are the initial times each object leaves. Choose t1 = 0 , then t2 = 1 (given). Also L=20 (given). And V is the initial velocity of object 2. This can be worked algebraically by solving for T in eq1 and substituting in eq2. Then solve for V. However, if you make the approximation g=10 (instead of 9.8) then it can be solved almost by inspection, without a calculater. Putting in the known values gives; 20 = 5T^2 20 = V(T-1) + 5(T-1)^2 From eq1, T=2 , so eq2 becomes; 20 = V + 5 , and so V = 15 which should be pretty close to the correct answer.
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∙ 15y agoanything shot up with that initial velocity. There isn't anything in specific.
convert degrees to meters
-- Acceleration of gravity, on or near Earth, is 9.8 meters ( 32.2 feet) per second2.-- Speed, neglecting the effects of air resistance, is9.8 meters (32.2 feet) per secondmultiplied by(number of seconds since the object was dropped)regardless of the mass, weight, or size of the object.
After 3.5 seconds of free-fall on or near the surface of the Earth, (ignoring effectsof air resistance), the vertical speed of an object starting from rest isg T = 3.5 g = 3.5 x 9.8 = 34.3 meters per second.With no initial horizontal component, the direction of such an object's velocitywhen it hits the ground is straight down.
D = 1/2 g T2T = sqrt(2D/g) = sqrt( 109.2 / 9.8 ) = 3.335 seconds(rounded)
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.
No, the acceleration is not the same for an object that is dropped and an object that is thrown. When an object is dropped, it experiences a constant acceleration due to gravity. When an object is thrown, its acceleration can vary depending on factors such as the initial velocity and direction.
If you simply release an object, the initial velocity is always zero.
It's not clear what you mean by the rate of the object, since objects don't have rates.When an object is dropped, on or near the Earth's surface, its rate of accelerationis 9.8 meters per second2, and its rate of speedincreases by 9.8 meters per secondevery second that it continues to fall.
4 Seconds
The object will be moving at 14.7 meters per second. 1.5 seconds X 9.8 meters per second squared(the gravitational constant). This assumes that the object's original velocity is zero.
An object dropped from near the Earth's surface will fall approximately 4.9 meters (16 feet) in the first second due to the acceleration of gravity. This distance is calculated using the formula s = 0.5 * g * t^2, where s is the distance, g is the acceleration due to gravity (9.8 m/s^2), and t is the time in seconds.
anything shot up with that initial velocity. There isn't anything in specific.
44 meters tall
The object will have the same acceleration of 9.8 meters per second squared whether you drop it or throw it downward. The initial velocity from throwing it will affect its overall velocity as it falls, but the acceleration due to gravity remains constant.
Yes, uniform negative acceleration (specifically gravity) can accurately describe the motion of a heavy object thrown downward from a tall building. The object would experience a constant acceleration due to gravity as it falls towards the ground. This acceleration would cause the object's velocity to increase over time until it reaches the ground.
The initial potential energy of an object dropped from a kilometer height is converted into both kinetic and potential energy as it falls. However, due to factors such as air resistance and sound produced during the fall, some of this energy is lost to the surroundings, resulting in a decrease in the final kinetic energy before impact compared to the initial potential energy.