Title: please explain what are the values of the verical and horizontal accerlrations of a projectile? Post by: rixter119 on Feb 5, 2013 please explain what are the values of the vertical and horizontal accelerations of a projectile?
i really dont understand this, please explain using easily understandable words THANKS! Title: please explain what are the values of the verical and horizontal accerlrations of a projectile? Post by: juli3_jon3s on Feb 5, 2013 Take your typical rifle say the AK 47. Then take your typical angle finder say the craftsman from Big Lots, and attach it to the barrel of your assault rifle using the magnet attached to the bottom. Then assume no gravity.
1)hold the gun so that the angle reads 0 and fire it. the bullet travels with 100% horizontal velocity and 0% vertical. expressed mathmatically (trigonometry) Sin = 0, Cos=1 2) repeat with the angle reading 90 and fire again. this time the bullet has 0% horizontal and 100% vertical velocity. Sin =1, cos =0. At any other elevation, there is a combination of vertical and horizontal components as the sine and cosine values lie between 0 &1. By multiplying the velocity by the sine for vertical component and the cosine for the horizontal, both components can be figured out. It is a basic right triangle solution. Vertical component is the rise and the horizontal is the run. The angle + the speed/velocity is called a vector and is indicated by an arrow at the angle of the barrel aka the angle of inclination . it gets a bit more complicated when we return to gravity because it imposes a downward component of -32 feet second. it is negative because it pulls the bullet down against the upward component of the bullet. of course if you shot down, then the vertical part would also be negative and gravity would increase the vertical component. hope this is understandable. Title: please explain what are the values of the verical and horizontal accerlrations of a projectile? Post by: jujublah on Feb 5, 2013 Think how a projectile is moving. It has upward and lateral displacements. To move up it should have certain velocity in the upward direction. This part of its velocity is called the vertical velocity. If Q is the angle made by the total velocity u with the horizontal, then
vertical velocity = usinQ To move laterally the projectile, it should have a velocity component in the that direction. We usually take this in a direction parallel to the ground and it is called the horizontal velocity. horizontal velocity = ucosQ Resultant velocity, u = square root of(hor. velocity^2 + vert. velocity^2). For more details. contact me. My id is rajsruthilayam@yahoo.co.in Title: please explain what are the values of the verical and horizontal accerlrations of a projectile? Post by: Bugbaja on Feb 5, 2013 Dearest B_Ball
============ Just an example for u OK ! -------------------------------- Physics/Math Question? --------------------------------- The launching velocity of a projectile is 20 m/s at 53 degrees above the horizontal. A. What is the vertical component of its velocity at launch? B. Its horizontal component of velocity? C. Neglecting air friction, which of these components remains constant throughout the flight path? D. Which of these components determines the projectile's time in the air? ------------------- just for reference g=32.2ft/s² r=unknown ?=0.5 * ((invsin)(r*g/v.²)) 2?=((invsin)(r*g/v.²)) 2?*v.²=((invsin)(r*g)) sin (2?*v.²)=r*g Vy=15.96 Vx=12.036 therefore sin (2?*v.²)/g=r and sin (2?*v.²)/r=g -------------------- first thing u do is look at the question of course u have 2 angles 1./ 53º and 2./ 90º-53º=37º you are given the resultant speed(hyp) so now u have to find the compotent speeds there are 2 speed(Y) and speed(X) so lets find both 53º above the horizon therefore 20 * cos(53º)=speed(X) or 20 * sin(37º)=speed(X) 20 * sin(53º)=speed(Y) or 20 * cos(37º)=speed(Y) so 20 * cos(53º)=speed(X) or 20 * sin(37º)=speed(X) 20 * 0.6018 =12.036 20 * sin(53º)=speed(Y) or 20 * sin(53º)=speed(Y) 20 * 0.798 = 15.96 now we have speed(Y) = 15.96 v= g * t v = 32.2 * 0.4956 v = 32.2 * 0.4956 v =15.96 or speed(Y) lets find distance to drop from the sky to ground d(Y) = 0.5 * g * t² d(Y) = 0.5 * 32.2 * (.4956)² d(Y) = 7.97ft or distance(Y) --------------------------- now u can calculate the Range ---------------------------- d(X) = speed(X) * t d(X) = 12.036 * .4956 d(X) = 5.965 ft Carry on from here now if u cant see this now take up social science or politics |