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a particle is projected vertically upward from ground with velocity 10 m/s. Find which time particle reaches highest point?
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A particle is projected vertically upward from ground with velocity 10 m/s. Find which time particle reaches highest point?
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at the max altitude, velocity is zero. Vf=vi-9.8t solve for t
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a particle is projected vertically upward with the velocity u m/s and after t(s) another particle is projected upwards from the
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Using t for time, and s for the delay, ut - (g/2)t^2 = u(t-s) - (g/2)(t-s)^2 ut - (g/2)t^2 = ut - us
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Stil wl b veloct of me nd -9.8m/s^2 for force of gravity
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To find the total time taken for the particle to hit the ground again, we can use the following
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a particle ia projected upward with a velocity of u m/s and after an interval of t(s) anoda particle is projected upward from
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To prove that the two particles will meet at a height of (4u^2/8g - g^2t^2/8g), let's break down the
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A partical is projected vertically upward with speed 20 m/s. What will be the time after particle is at a height 15m above the
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h=ho+vi*t-1/2 g t^2 15=0+20t-4.9t^2 Put it in the form of a quadratic equation at^2+bt+c=0 and use
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A particle of mass 3.00 kg is projected vertically upward with an initial velocity of 20.0 m/s. Neglecting air resistance, the
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i need help
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A particle is projected horizontally with velocity V m/s, from a point h meters above ground. Taking g ms^-2 as the acceleration
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To show the equation of the path, we can use the equations of motion for vertical and horizontal
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I'm having a bit of trouble with F. Force is in units of N = kg-m/s^2 Having v^2 there makes the
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A particle A is projected vertically upwards from a point O on horizontal ground with speed 20ms. At the same instant a particle
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To find the time elapsed between B's arrival on the ground and A's arrival on the ground, we first
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