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# Answer to Question #117297 in Mechanics | Relativity for Meryclare

Question #117297
A particle is projected at an angle of 60O to the horizontal with a speed of 20 m/s and total time of pacicle flight at 3.46s Calculate of Speed of the particle at its maximum height [take g = 10 m/s2 ]
1
2020-05-20T09:30:18-0400

We have given,

Angle of projection "\\theta=60^{\\circ}" ,initial speed "u=20m\/s" ,Time of flight "T_{flight}=3.46s" .

We have to calculate the speed at maximum height "(H)" .

Since, we know that height "H" is achieved when the "y" component of the final velocity becomes at time

"t=\\frac{1}{2}T_{flight}"

But we noticed that "x" component of the velocity of the particle does not change with time as there is no horizontal force acting on the particle, hence no acceleration i.e "a_x=0"

Thus, there is only horizontal component of velocity will survive, hence from the given data,

"v_y=0"

"v_x=u\\cos(\\theta)-a_xt\\\\\n\\implies v_x=u\\cos(\\theta)\\\\\n\\implies v_x=20\\cdot \\cos(60^{\\circ})=10ms^{-1}"

Therefore, velocity of the particle at maximum height is "10ms^{-1}"

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