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?

1 Answer
Dec 23, 2015

After 0.094"s"

Explanation:

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"