A projectile is fixed with a velocity of 20 m/s at an angle of 400...

A projectile is fixed with a velocity of 20 m/s at an angle of 400 to the horizontal. Determine the components of the velocity of the projectile at its maximum height.?

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Answers (2)

Jace
1 week ago
Maximum height = u^2*sin^2teta/2g
=20^2*sin^2(400)/2*10
=400*(0.6)^2/20
=400*0.36/20
=144/20
=7.2m
so the velocity
v^2=u^2+2as
v^2=2*10*7.2
v^2=144
v=12m/s
mustapha
1 week ago
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