Ignoring drag:
h=32.7*t-0.5*g*t2
Where g is about equal to 32.2 ft/sec2
it means initial upwards height times time in seconds
The equation for vertical motion is y = v0t + .5at2. y is vertical displacement v0 is initial vertical velocity a is acceleration (in meters, normal gravitational acceleration is about -9.8 m/s/s, assuming positive y is upward displacement and negative y is downward displacement)
The baseball will travel upwards until its velocity is zero and then fall back down again. The acceleration of the baseball is the constant acceleration due to gravity acting in a downwards direction. The time taken to fall back down will be the same as the time to climb, thus the total time is twice the climb time. Use Netwon's equations of motion and ignore air resistance, the time to the top of the climb is: v = u + at → t = v - u/a v = final velocity = 0 m/s at the top of the climb u = initial velocity = +100 m/s a = acceleration = -g m/s² t_top = 0 - (100 m/s) / (-g m/s²) = 100/g s → t_total = 2 × t_top = 2 × 100/g s = 200/g s ≈ 200/9.81 s ≈ 20.4 s
There's no such thing as "time of the downward velocity", but I think I get the sense of your question. If the effects of air resistance can be disregarded, then any object thrown upwards spends half of its time rising, and the identical amount of time falling back to the height of your hand when you let it go.
maximum velocity is the highest possibly speed an object can travel before the forces acting on it reach an equilibrium and it is no longer able to accelerate. For example a parachutist will fall and accelerate rapidly until the air resistance pushing upwards against her downward force becomes balanced and her speed is steady, its more commonly known as 'terminal velocity' not maximum.
it means initial upwards height times time in seconds
When an object is moving upwards, its velocity is directed upwards. If the object is near the Earth or any other planet, then its acceleration is directed downwards, which also means that its upward velocity is decreasing.
In that case, the velocity is zero.
Using the equation: x=vot+(1/2)at2 x: -39.2m (though the object is thrown upwards, its total displacement is just the amount it fell from the tower). vo: initial velocity t: 4 s a: -9.8m/s2 (assuming we're on the earth's surface) -39.2=vo(4 s)+(1/2) (-9.8m/s2)(4 s)2 ((78.4 m - 39.2 m)/ 4 s) = v0 So the initial velocity is 9.8 m/s.
It DOES NOT stop for a 'few seconds' . It stop only for an instant as its movement changes from upwards to downwards.
The velocity of such an object changes all the time. Assuming you throw something directly upwards and there is no wind, it will go upwards, slower and slower, until it reaches its highest point. At that moment, its velocity is zero. Then, still as a result of gravity, it will move downward, faster and faster.
when a ball is thrown upwards velocity increases but acceleration decreases hence making it anti parallel to each other
* the line or plane indicating the limit or extent of something * a light, self-propelled movement upwards or forwards
An object can be thrown vertically upwards or at an angle to the ground, in both cases it is needed that time of flight be 5 seconds. This means it's time of ascent (going up) is 2.5 seconds and time of descent(coming down) is also 2.5 seconds. So, it reaches highest point 2.5 seconds after it is thrown. At highest point the vertical component of velocity1 of the object becomes zero for an instance.Now, kinematics equation can be used to solve this.v = uy - g*twhere v is final velocity at top =0. uy is initial vertical velocity1. g is accleration due to gravity(9.8ms-2). t is time of ascent.putting values we get.0 = uy - 9.8 * 2.5or uy = 24.5 ms-1So we need to throw an object with vertical velocity = 24.5 meters per second so that it remains in air for 5 seconds.1. If object is thrown at an angle then vertical component of velocity of projection is taken.If object is thrown vertically upwards then vertical component of velocity of projection issame as velocity.vertical component of velocity of projection(uy) = u*sin(θ), where u is velocity of projectionand θ is angle of projection with respect to horizontal.2. Accleration due to gravity is different for different places. 9.8ms-2 is an approximatevalue.3. Here, air resistance and wind speeds have been neglected as they make the calculation verytedious and they are always varying from time to time and place to place.
Its velocity decreases because gravity is pulling on it as it goes up. Its acceleration increases.
Acceleration occurs when velocity changes over time. The formula for it is as follows: a = (Vf - Vi) / t a: acceleration (meters/seconds2) Vf: Final velocity (meters/seconds) Vi: Initial Velocity (meters/seconds) t: Time (seconds)
It launches a a rocket vertically upwards with an initial speed of 40.0 m/s.