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a projectile is fired with a velocity of 320 m/s at an angle of 30 to the horizontal. find the time to reach it's greatest height, |
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Answer» Explanation: Given: angle θ = 30˚ ; Initial velocity = 20m/s Solve for maximum height dymax = ? There are three equations to do this. The first EQUATION is solving for the initial vertical velocity; the second equation is solving for the time of RISE and the third equation is solving for the maximum height. Solving for the initial vertical velocity : initial vertical velocity = initial velocity * SIN 30 initial vertical velocity = 20 m/s * 0.5 initial vertical velocity = 10 m/s Solving for the time of rise time_up = initial vertical velocity / acceleration due to gravity time_up = 10 m/s / 9.8 m/s^2 time_up = 1.0204 seconds Solving for the maximum vertical height of rise dymax = Viy * t – 1/2 gt^2 dymax = 10 m/s * 1.0204 s – 4.9 m/s^2 * (1.0204 s)^2 dymax = 10.204 m – 4.9 * 1.0412 m dymax = 10.204 m – 5.10196 m dymax = 5.10 m The body rises up to a maximum height of 5.10 METERS. |
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