Perhaps you meant speed or velocity, because the vertical acceleration is constant throughout the bomb's decent, ignoring the effects of air resistance. The acceleration due to gravity is 9.8 m/s2 for all values of t.
Distance of fall in T seconds = 1/2 g T2Distance of fall in 2 seconds = (1/2) (9.8) (2)2 = (4.9 x 4) = 19.6 metersHeight of this particular ball after 2 seconds = (70 - 19.6) = 50.4 meters
distance (s) = 1/2 acceleration (a) x time (t) squared or s = 1/2 at^2.; a = gravity acceleration = 9.81m/s/s so time is 1.1 seconds
No, acceleration due to gravity is a constant at 9.81ms-2. It cannot be influenced by other factors such as height.
Ep (joules) = mass * acceleration due to gravity * height So: height = Ep / (mass * acceleration due to gravity)
The effect of increasing the height of the track on the acceleration of the object is that more work is required to accelerate. It increases the gravity.
Its acceleration is always the same - the acceleration of gravity at 32 ft/sec/sec - no matter what distance it is during drop, until it hits the ground.
Assuming that seconds refers to the period, the frequency is the reciprocal (1 / period in seconds). The height of the wave is irrelevant in this case.
Regardless of the height from which it is falling, (neglecting air resistance) it's speed will be 19.62 metres per second. (Acceleration from gravity is 9.81 metres per second squared, so after 1 second it is moving at 9.81 metres per second and after 2 seconds it is moving at 19.62 metres per second.
secret
mgh represents the potential energy of an object located at a height h above the ground, where m is the mass of the object, g is the acceleration due to gravity, and h is the height. It is calculated as the product of the mass, acceleration due to gravity, and the height.
Acceleration due to the force of gravity.
acceleration