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If the object is moving upwards with a constant velocity, there is only one arrow in the diagram, and it points straight down, due to gravity. If it has a force pulling upward on it, there are two arrows, one up, due to the force, and one down, again, due to gravity.
If you ignore the effect of the air grabbing at it and only figure in gravity, then the horizontal component of velocity is constant, from the time the stone leaves your hand until the time it hits the ground. Makes no difference whether you toss it up, down, horizontal, or on a slant. Also makes no difference whether it's a cannonball, a stone, or a bullet.
When we throw the object upwards we consider that upward direction as positive. Therefore, the velocity in that direction is positive but the acceleration due to gravity is in the opposite direction and so it is considered negative. But when the ball comes down again after reaching a certain height the velocity is in opposite direction to the earlier one and so the velocity now is negative as a result the acceleration is again negative.
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.
easy for upward path applying eqn v^2 - u^2 = 2aS and v =0, a = -g = -9.8 we get s = 11.479 now on way down it was cought on 2.5 m above ground so s for downward = 9.479 so as u for downward = 0, a = g = 9.8 , s = 9.479 applying v^2 - u^2 = 2aS again we get v = 13.63 mps now as v = u +at therf. t = 1.39 :) by ravinder balhara. :)
Its recycled again;D
By the end keeping the planes of the boomerang basicaly horisontal to the ground, again and again until you have it right,
Yes he will
get good deals
Return means to come back again.
because they sell for less..
If it was thrown hard enough. Than again, pretty much any solid object can shatter a window if thrown hard enough...
never again.
If the object is moving upwards with a constant velocity, there is only one arrow in the diagram, and it points straight down, due to gravity. If it has a force pulling upward on it, there are two arrows, one up, due to the force, and one down, again, due to gravity.
If you ignore the effect of the air grabbing at it and only figure in gravity, then the horizontal component of velocity is constant, from the time the stone leaves your hand until the time it hits the ground. Makes no difference whether you toss it up, down, horizontal, or on a slant. Also makes no difference whether it's a cannonball, a stone, or a bullet.
Percy created a air bubble and then they kissed again.
Nothing has been confirmed for his return. (April 2010)