Projectile DynamicsDate: 10/21/2000 at 08:20:04 From: Peter Subject: Dynamics A particle is projected in a parabolic path from a point on a flat surface. If the angle of trajectory is theta and the initial velocity is v, show that at time t when the angle to the horizontal is now gamma: tan(gamma) = tan(theta) - (gt/v) sec(theta) Thank you. Any help would be appreciated. Date: 10/21/2000 at 11:30:51 From: Doctor Robert Subject: Re: Dynamics This problem is not as hard as it seems. Think about the velocity vector of the projectile as the projectile goes along its trajectory. First of all, if you neglect air resistance, there is no change in the horizontal component of the velocity vector; it's always v cos(theta). Now the vertical component of the velocity vector is v sin(theta) - gt where g is the acceleration of gravity and t is the time of flight. At some time the vertical component becomes zero; this is when the projectile is at the top of its trajectory. Now, draw a diagram of the velocity vector and show the components. The angle that the velocity vector makes with the horizontal is gamma. So from your diagram, tan(gamma) = [v sin(theta)-gt]/[v cos(theta)] If you simplify this just a little, it gives you the desired result. - Doctor Robert, The Math Forum http://mathforum.org/dr.math/ |
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