A pellet launch is found to propel a metal ball sideways a distance of 2m, when the angle of launch is 36.9 degrees.
a) What do you have to do to the initial velocity- in terms of x and y components?
b) What variable is common to both horizontal and verticle motion? Find the variable.
c) Use the b) value to answer this: If the muzzle velocity is 5m/s, how far above or below the starting point does the ball hit the ground?
a) You need to write the initial velocity as a vector. This vector has a magnitue of 5.00 m/s and a direction of 36.9 degrees relative to the horizon. Therefore, the horizontal velocity is 5.00*cos(36.9)=4.00 m/s, and the vertical velocity is 5.00*sin(36.9)=3.00 m/s in the upward direction. These are the x and y components.
b) We have enough ...
This answers the vertical flight distance for a launched pellet with a specified angle and sideways distance. The velocity and variable of time is looked at to determine the solution.