The ball continues to rise until its upward speed decreases to zero. Since gravity adds 9.8 meters per second
to its downward speed every second, it will continue moving upward for (20/9.8) = 2.0408 seconds (rounded).
After it peaks and begins falling, it will return to the elevation from which it was launched in the same length of time.
The total time from launch to catch is (20/9.8) x 2 = 4.0816 seconds (rounded).
(We don't know about additional time to fall to the ground, because we don't know exactly how high
the hand was when the ball was launched.)
2.08 seconds
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.
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.
After just over three and a quarter seconds.
The maximum height of a thrown ball is dependent on the upward portion of the initial velocity. Air friction will slow it somewhat but gravity will cause it to lose most of its upward velocity. The velocity will decrease by 9.8 m/sec for each second of its upward motion, until it reaches zero. At that point, the ball is pulled back toward Earth.
About 11 miles per hour.
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.
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.
After just over three and a quarter seconds.
anything shot up with that initial velocity. There isn't anything in specific.
The maximum height of a thrown ball is dependent on the upward portion of the initial velocity. Air friction will slow it somewhat but gravity will cause it to lose most of its upward velocity. The velocity will decrease by 9.8 m/sec for each second of its upward motion, until it reaches zero. At that point, the ball is pulled back toward Earth.
About 11 miles per hour.
1 sec : position = 10.1 metres above your hand, velocity = 5.2 ms^-1.40 sec : position = 7240 metres below your hand, velocity = 377 ms^-1 downwards.
By the Way, guys, this is based on the equation H= -16t2+vt+s
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.
0.82 metres.
0.82 metres.
its upward at some specified angle