After 3.5 seconds of free-fall on or near the surface of the Earth, (ignoring effects
of air resistance), the vertical speed of an object starting from rest is
g 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 velocity
when it hits the ground is straight down.
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.
Let v be the velocity when the ball is at 640 feets going downwards v = 48 feet /sec let the velocity with which it reaches the ground be u then, u2=v2+2gh g = acc due t ogravity in feet/sq.sec h = 640 feet the time taken to reach the ground = time to return to 640ft + the time to fall from there Time taken to get to the ground is 8 seconds. Final velocity is 208 feet per second downwards
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.
A child drops a ball from a window. The ball strikes the ground in 3.0 seconds. What is the velocity of the ball the instant before it hits the ground?
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.
39 m\s downward
The speed is 44.4... repeating metres per second.
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.
They will both have the same speed because the will hit the ground at the same time, due to vertical velocity.
Simply use the expression v = gt g = 9.8 m/s^2 and t given as 4.5 s So velocity with which the penny hits the ground will be 44.1 m/s
If it was thrown horizontally or dropped, and hit the ground 3.03 seconds later, then it hit the ground moving at a speed of 29.694 meters (97.42-ft) per second. If it was tossed at any angle not horizontal, and hit the ground 3.03 seconds later, we need to know the direction it was launched, in order to calculate the speed with which it hit the ground.
The initial velocity of a dropped ball is zero in the y (up-down) direction. After it is dropped gravity causes an acceleration, which causes the velocity to increase. F = ma, The acceleration due to gravity creates a force on the mass of the ball.
Let v be the velocity when the ball is at 640 feets going downwards v = 48 feet /sec let the velocity with which it reaches the ground be u then, u2=v2+2gh g = acc due t ogravity in feet/sq.sec h = 640 feet the time taken to reach the ground = time to return to 640ft + the time to fall from there Time taken to get to the ground is 8 seconds. Final velocity is 208 feet per second downwards
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.
The volleyball will NOT hit the ground with greater anything. Assuming that the soccer ball is the same spherical diameter and greater mass than the volleyball it will hit the ground with greater velocity and greater impact.
Disregarding air resistance, what is the speed of a ball dropped from 12 feet just before it hits the ground? (Use 1 ft = 0.30 m, and use g = 9.8 m/s2.)
31 m/s