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

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Answer 1 · 2015-12-23T17:24:43

After $0.094 s$ $0.094"s"$

The vertical component of the velocity is given by:

$v_{y} = v sin θ$ $v_y=vsintheta$

Use the equation of motion:

$v = u + a t$ $v=u+at$

This becomes:

$v = u - g t$ $v=u-"g"t$

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

$:.t=(vsintheta)/g$

Convert from radians to degrees:

$2pi=360^@$

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

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

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

If a projectile is shot at an angle of (3pi)/83π8(3pi)/8 and at a velocity of 1 m/s1ms1 m/s, when will it reach its maximum height?