Juan throws a baseball with an initial speed of 42 m/s at an angle of 45 0 above the horizontal. Richard catches the ball at the same height that Juan threw it. How far did Juan throw the ball?

Answer 1

This kind of projectile motion can be decomposed into two separate motion:

1. A constant velocity motion in the horizontal direction, or x direction, at a speed of V*cos(theta)
2. A constant acceleration motion in the vertical direction, or y direction, at a initial speed of v*sin(theta)
3. The time the ball flies in the air can be decided by the following fact: By the time the balls falls back down to the same height, the displacement on the y direction is 0, thus

v*sin(theta)*t - 1/2gt² = 0, thus t = 2v*sin(theta)/g

1. The displacement on the x direction can then be found by

dx = v*cos(theta)*t = v*cos(ethta) * 2v*sin(theta)/g = 2v²sin(theta)*cos(theta)/g

= v²sin(2*theta)/g

1. Note v = 42, theta = 45, and g = 9.8, so we get

dx = 42² * 1.0 /9.8 = 180m

1. Note from trigonometry, sin(2*theta) = 2*sin(theta) * cos(theta)