If a projectile is shot at an angle of $\frac{2 π}{3}$ $(2pi)/3$ and at a velocity of $23 \frac{m}{s}$ $23 m/s$ , when will it reach its maximum height?

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Answer 1 · 2016-05-02T11:41:23

$t = 2, 03 s$ $t=2,03 " s"$

$you can use the formula t = \frac{v_{i} \cdot sin α}{g}$ $" you can use the formula "t=(v_i*sin alpha)/g$

$v_{i} = 23 \frac{m}{s}$ $v_i=23 m/s$

$α = \frac{2 π}{3}$ $alpha=(2pi)/3$

$g = 9, 81$ $g=9,81$

$t = \frac{23 \cdot sin (\frac{2 π}{3})}{9, 81}$ $t=(23*sin((2pi)/3))/(9,81)$

$t = 2, 03 s$ $t=2,03 " s"$

If a projectile is shot at an angle of (2pi)/32π3(2pi)/3 and at a velocity of 23 m/s23ms23 m/s, when will it reach its maximum height?