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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.
Zero
velocity is found by dividing the distance with time. In a second the height traveled is found by multiplying the velocity by the time taken and then dividing the answer by two.
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 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.
when the object reaches maximum height, the velocity of the object is 0 m/s.It reaches maximum height when the gravity of the body has slowed its velocity to 0 m/s. If there is no gravity and there is no external force acting on it then it will never reach a maximum height as there wont be a negativeaccelerationdemonstrated by newtons first law.Where there is and you have the objects initial velocity then you can use :v^2 = u^2+2.a.sv = Velocity when it reaches Max. height so v = 0u = Initial Velocity (m/s)a = Retardation/ Negative Acceleration due to gravity, -9.80m/s ^2And then the unknown (s) is the displacement, or height above ground, and if everything else is in the right format it should be in metres.
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.
Zero
Height reached = 3.7 metres.The mass of the ball is not really relevant.
velocity is found by dividing the distance with time. In a second the height traveled is found by multiplying the velocity by the time taken and then dividing the answer by two.
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.
If you ignore air resistance, then they will reach their maximum height at the same time. In order not to ignore air resistance, you would need to know their shapes.
30 mph!
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.
whenever an object is thrown in the air we must know the initial velocity with which the object has been thrown.
A stone is thrown with an angle of 530 to the horizontal with an initial velocity of 20 m/s, assume g=10 m/s2. Calculate: a) The time it will stay in the air? b) How far will the stone travel before it hits the ground (the range)? c) What will be the maximum height the stone will reach?
A projectile that is thrown with an initial velocity,that has a horizontal component of 4 m/s, its horizontal speed after 3s will still be 4m/s.