If a projectile is shot at an angle of $\frac{5 π}{12}$ $(5pi)/12$ 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 · 2018-07-09T22:04:57

$0.04755416421469008\ m$

The maximum height $H_{max}$ obtained by a projectile shot at an angle $\theta={5\pi}/12$ with initial velocity $u=1\ m/s$ is given as follows

$v^2=u^2+2as$ (from II equation of motion)

$0^2=(u\sin\theta)^2-2gH_{max}$

$H_{max}=\frac{u^2\sin\theta}{2g}$

$H_{max}=\frac{1^2\sin^2({5\pi}/12)}{2\times 9.81}$

$=0.04755416421469008\ m$

If a projectile is shot at an angle of (5pi)/125π12(5pi)/12 and at a velocity of 1 m/s1ms1 m/s, when will it reach its maximum height?