If a projectile is shot at an angle of $\frac{π}{3}$ $pi/3$ and at a velocity of $6 \frac{m}{s}$ $6 m/s$ , when will it reach its maximum height??

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If a projectile is shot at an angle of $\frac{π}{3}$ $pi/3$ and at a velocity of $6 \frac{m}{s}$ $6 m/s$ , when will it reach its maximum height??

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Answer 1 · 2016-02-15T18:43:27

$v_{f} = v_{i} + a t; a = - g; v_{f} = 0$ $v_f = v_i + at; a=-g; v_f = 0$
$0 = 3 \sqrt{3} - 9.81 t; t = 3 \frac{\sqrt{3}}{9.81} \approx .53 s$ $0 = 3sqrt(3) - 9.81t; t=3sqrt(3)/9.81 ~~ .53 s$

Explanation:

GIVEN: a projectile shot at angle, $θ = \frac{π}{3}; v_{i} = 6 \frac{m}{s}$ $theta = pi/3; v_i = 6m/s$
find the time require maximum height, i.e. zero velocity at tyhe top.
DEFINITION AND PRINCIPLES: here what tou need a kinematic equation that use initial velocity, final velocity and acceleration (gravity in this case)
$v_{f_{y}} = v_{i_{y}} + a_{y} t; a = - g; v_{f_{y}} = 0; v_{i_{y}} = 6 sin (\frac{π}{3}) = 3 \sqrt{3};$ $v_(f_y) = v_(i_y) + a_yt; a=-g; v_(f_y)= 0; v_(i_y)=6 sin(pi/3)=3sqrt(3);$
$0 = 3 \sqrt{3} - 9.81 t; t = 3 \frac{\sqrt{3}}{9.81} \approx .53 s$ $0 = 3sqrt(3) - 9.81t; t=3sqrt(3)/9.81 ~~ .53 s$

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If a projectile is shot at an angle of $\frac{π}{3}$ $pi/3$ and at a velocity of $6 \frac{m}{s}$ $6 m/s$ , when will it reach its maximum height??

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Explanation:

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If a projectile is shot at an angle of $\frac{π}{3}$ $pi/3$ and at a velocity of $6 \frac{m}{s}$ $6 m/s$ , when will it reach its maximum height??