How to explain maximum range to a child? How would I explain that the maximum range of a projectile occurs when the projectile is launched at $\theta = 45^{\circ}$ to a child?
I realize that I’ve never conceptualized why the maximum range would occur at this angle. I have only seen derivations going from the kinematic equations, but that would only confuse a child. Is there an intuitive explanation?
 A: Ask the child to imagine the two limiting cases:

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*The projectile is launched at an angle of 0° --- perfectly horizontally. The projectile will immediately hit the floor and stop.


*The projectile is launched at an angle of 90° --- perfectly vertically. The projectile will fly high upwards, stay a good bit in the air, and then fall back down. But it will fall exactly where it already was.
In both cases, the projectile doesn't move horizontally. But we obviously know that if the projectile is launched between those two extremes, it's going to move something. So there must be some intermediate angle where we reach the maximum distance (there could be more than one that would give this maximum distance). We want to balance how long the projectile stays in the air with how fast it's going horizontally to achieve that maximum distance. It so happens that that angle is 45°, smack in the middle. Any higher and the projectile will slow down too much, and the extra time it spends in the air won't be enough to balance that out; any lower and it won't stay in the air for long enough, and its extra horizontal speed won't last long enough to make a positive difference.
I'm not sure how you could prove this without any math though.
A: In addition to the jvf answer, if the child insists to ask why the maximum is exactly at $45^{\circ}$, it is possible to use that mathematically the situation is the same as to get the largest area of a rectangle, keeping the same diagonal and changing the sides. It is intuitive to realize that the maximum area is when the rectangle turns on a square.
The diagonal is the magnitude of the launch speed, and one of the angles of the sides with the diagonal is the launch angle.
A: It all depends on the age of the child so even the equation distance traveled = speed $\times$ time might be conceptually too difficult.  The outline that follows cab be expanded on as one feels fit by use of the Phet simulation mentioned at the end might make the subsequent explanation much better and more interesting in that the $45^\circ$ maximum range can be found by "experiment".
The range is the horizontal distance traveled by an object before it hits the ground.
You might start off by saying that when you throw an object at some angle relative to the ground you can think of the motion of the object as being made up of two speed, a horizontal speed and a vertical speed. Note that I would not use the word velocity.
If there is no air resistance then the horizontal speed does not change whereas the vertical speed decreases to zero as the object is going up, it then stops at its greatest height and then start to come down with increasing downward speed.
The horizontal the object travels before hitting the ground depends on the horizontal speed and the time the object is in the air.
The time the object is in the air depends on the vertical speed.
You can maximize the time the object spends in the air by throwing it vertically upwards which unfortunately means that the object has no horizontal speed so the range is zero, the object hits the ground at the point where it was thrown upwards.
So you compromise a little by throwing the object at an angle of $80^\circ$.
The object does not stay up in the air as long because the vertical speed is now less but is does have a small horizontal speed and so hits the ground away from where the object is thrown.
Decreasing the angle at which the object is thrown does decrease the time the object is in the air but as the horizontal speed is now larger the range will also be larger.
The nest one can do is to make the vertical and horizontal speeds the same ie throw the object at an angle of $45^\circ$.  In such a case the time the object is in the air is less than for angles greater than $45^\circ$ but this is compensated for by the greater value of the horizontal speed.
Smaller angle have a greater horizontal speed but less time for the object to be in the air and the range is smaller than for $45^\circ$.
And rather than do the explaining just verbally or with pencil and paper try use the Phet simulation Projectile Motion - the Intro and with the cannon at zero height.
