The 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.
anything shot up with that initial velocity. There isn't anything in specific.
-- 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)
That's the formula for the height of an object that was tossed upward at a speed of 40 meters per second, after ' t ' seconds . This object has to be something like a canonball, or a baseball pitched by a professional etc. The initial vertical speed of 40 meters per second is almost 90 miles per hour upward !
The acceleration of an object dropped from a height of 10 meters is approximately 9.81 m/s2.
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.
anything shot up with that initial velocity. There isn't anything in specific.
44 meters tall
Whether the object is dropped, thrown downwards, thrown upwards, or thrown horizontally, its downward acceleration is the same 9.8 meters per second2. If it's thrown downwards, however, its speed at any instant is greater than the speed at the same instant would be if it had only been dropped, since it has some speed before the acceleration begins.
The total displacement is 30 meters South. Displacement is the difference between the initial and final positions of an object, irrespective of the path taken. In this case, the person returns partially to the initial position after moving North by 50 meters.
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.