Want this question answered?
less than the speed it had when thrown upward.
It comes back downward! :) enjoi!
zero
If a ball is thrown vertically upward with a velocity of 160 ft/s, then its height after t seconds is s = 160t - 16t^2. If a ball is thrown vertically upward with a velocity of 160 ft/s, then its height after t seconds is s = 160t − 16t^2.
The speed/force which was imparted to it when it left the thrower's hand.
because there if speed is constant than ball never come back to earth hence speed of ball is not constant
less than the speed it had when thrown upward.
zero
It comes back downward! :) enjoi!
If a ball is thrown vertically upward with a velocity of 160 ft/s, then its height after t seconds is s = 160t - 16t^2. If a ball is thrown vertically upward with a velocity of 160 ft/s, then its height after t seconds is s = 160t − 16t^2.
The speed/force which was imparted to it when it left the thrower's hand.
About 11 miles per hour.
t matters how much mass the ball has
If a ball is thrown horizontally from a window on the second floor of a building, the vertical component of its initial velocity is zero.
The thrown ball will (usually) have the highest velocity as the acceleration (resultant of force) used to throw it exceeds that of the other two balls. The ball thrown upward will have a higher downward velocity than the dropped ball even though their accelerations (due to gravity) are the same, as it has more time to travel downward. Although, If the ball thrown upward is thrown high enough, it may even travel faster than the ball thrown downward if the downward throw's force is not enough to beat the ball's terminal velocity (quite a bit of height would be required though).
The ball can be considered a closed system.
A ball thrown vertically upward returns to the starting point in 8 seconds.-- Its velocity was upward for 4 seconds and downward for the other 4 seconds.-- Its velocity was zero at the turning point, exactly 4 seconds after leaving the hand.-- During the first 4 seconds, gravitational acceleration reduced the magnitude of its upward velocity by(9.8 meters/second2) x (4 seconds) = 39.2 meters per second-- So that had to be the magnitude of its initial upward velocity.