Launch angles and height of launch

Question

When an object is launched at ground level, the angle of launch that will make the object go the furthest is 45. If we were to change the height of the initial launch to 10 m, why is the launch angle of 35 degrees launch the object further than 45 degrees? (with the same initial velocity)

Answer 1

The general rule for projectile angles is that at 45 degrees you get the maximum horizontal displacement because you have the maximum horizontal speed component combined with the largest air time possible at that speed ,

< 45 degrees means less time in the air to travel horizontally at the less than maximum horizontal speed component possible of the launch speed so there is less displacement horizontally.

> 45 degrees means that the object has more time in the air but the horizontal component of the launch speed is lower so there is less horizontal displacement.

If you launch a projectile 10 meters above the ground at 35 degrees , that means you have more horizontal speed with about 0.3 seconds less total air time(assuming 10 m/s launch speed) to reach the height you launched from BUT you have 1 more second air(g = 10 m/s²) time to travel to the ground horizontally at that 35 degree angle below the launch level.

Using a launch velocity of 10 m/s and solving for horizontal displacement at 45 degrees and 35 degrees, the object travels 9.9 m and 9.3 m respectively to reach the launch levels. In the end, if you solve for the horizontal displacement traveled in the one second remaining after the second object passes the launch level( using trig functions at 35 degree angle solve for d_x, d_y = 10 m), the extra horizontal distance traveled in that one second at the horizontal component speed of 8.2 m/s = 8.2 m. If you add this horizontal 8.2 m to the 9.3 m already traveled to get back at launch level, the object travels 17.5 m total

Answer 2

The first answer is correct. I've graphed his answer, using his values for the initial velocity and acceleration of gravity.

Answer 3

Edgar:

Simple answer: that statement is false!

The maximum range is always from a launch angle (LA) of 45^o.

Look at it this way: imagine a horizontal surface parallel to the ground and 10 m above ground.

Then, you will agree the maximum range is at LA = 45^o.

Now, see that the projectile, after it reaches its maximum range on the 10m surface, continues downward until it strikes the ground, 10 m below. It will still be further than any other other LA.