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August 29, 2014

August 29, 2014

Posted by **Candice** on Saturday, July 14, 2012 at 4:38am.

Find the speed and direction of a particle which, when projected from a point 15 m above the horizontal ground, just clears the top of a wall 26.25 m high and 30 m away.

Thanks in advance

- Math -
**Henry**, Monday, July 16, 2012 at 4:33pmh1 = 15 m.

h2 = 26.25 - 15 = 11.25 m. = Ver. dist.

from firing point to top of wall.

Y = h2 = 11.25 m.

X = 30 m.

tanA = Y/X = 11.25 / 30 = 0.375

A = 20.6 Deg. = Direction of particle.

Range = Vo^2*sin(2A)/g = 30 m.

Vo^2*sin(41.2) / 9.8 = 30

Divide both sides by sin(41.2):

Vo^2 / 9.8 = 45.5

Vo^2 = 446.3

Vo = 21.1 m/s @ 20.6 Deg. = Velocity

and direction.

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