If a projectile is shot at an angle of #(3pi)/8# and at a velocity of #1 m/s#, when will it reach its maximum height?

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Answer 1 · 2015-12-23T17:24:43

After #0.094"s"#

The vertical component of the velocity is given by:

#v_y=vsintheta#

Use the equation of motion:

#v=u+at#

This becomes:

#v=u-"g"t#

#:.0=vsintheta-"g"t#

#:.t=(vsintheta)/g#

Convert from radians to degrees:

#2pi=360^@#

#:.pi=360/2=180^@#

#:.(3pi)/8=(3xx180)/8=67.5^@=theta#

#:.t=(1xx0.923)/9.8=0.094"s"#