After just over three and a quarter 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.
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.
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.
its upward at some specified angle
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.
it would be slower
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.
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.
its upward at some specified angle
anything shot up with that initial velocity. There isn't anything in specific.
About 11 miles per hour.
the initial velocity of the rocket is zero.
The kinetic energy will be decrease because when the body thrown in upward direction the gravitation force will be act in opposite direction which reduce the velocity.so the velocity will decrease.
Ignoring air resistance, I get this formula:Maximum height of a vertically-launched object = 1.5 square of initial speed/GI could be wrong. In that case, the unused portion of my fee will be cheerfully refunded.
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.