At what angle should we launch a projectile in order to maximize its horizontal range?

Question

Now calculate the range of a ball in yards hit at this launch angle with an initial speed of 90 miles per hour.

Answer 1

You must know that, assuming no air resistance, the angle of projection for maximum range with a given initial speed is 45 degrees to the horizontal (or vertical, for that matter)! (The problem is much more interesting if some particular law of resistance is posited.)

With that angle of projection, of course, if the speed of projection is v, both the VERTICAL, and the HORIZONTAL components of VELOCITY are both v_0 = v / [sqrt(2)] or (1/2) v sqrt(2) or 0.707107 v (to 6 significant figures). From there, it is easy to work out:

(a) The time of flight, and

(b) The range with that time.

So, (a) the vertical velocity at any stage is v_v = v_0 - g t. You can either say that by symmetry, the time of flight T is TWICE the time to give v_v = 0, or alternatively it's the time to give you v_v = - v_0. By either of these arguments, T = 2 v_0 / g.

Then (b,) the RANGE = v_0 x T = 2 (v_0)^2 / g; but (v_0)^2 = (1/2) v^2. Therefore the maximum range is given by:

v^2 / g ! (What could be nicer?)

Now all we have to do is convert v (in mph) to feet per second, to have compatible units with the usual expression for g, 32.2 ft / sec^2.

Here, one's State Drivers' Handbook comes in useful: it tells you that 60 mph is 88 ft per sec. So 90 mph = 132 ft per sec. Hence the maximum range with a projection speed of 90 mph is:

(132)^2 / 32.2 feet = 541.118 feet, to 6 significant figures.

This seems very reasonable; it's about 180 yards.***

Live long and prosper.

P.S. I have to admit that I chortled at the idea of it being 810 miles, as someone else deduced! We need a little reality check here: good cricket bowlers, baseball pitchers, batters and tennis players are all capable of throwing/hitting balls at speeds on the order of 100 mph. Does anyone seriously think that those balls could then fly downrange for anything remotely approaching 1000 miles ?! There is a point where, even in science, common sense should intervene to tell you that you must have made a mistake somewhere. The UNITS in that particular calculation were of course multiply incompatible --- with a speed in miles per hour, and the gravitational acceleration in metres per second squared. An absurd conclusion like that should always send one back to check the calculation carefully.

Answer 2

Use 45 degrees for your calculations. Should work out something like this. At 45 degrees, half the velocity is vertical and half horizontal. You can calculate the flight time by working out the time for the ball to go from zero altitude to its maximum altitude, using half of the velocity for the vertical component. Knowing the flight time, you can crank out the horizontal distance.

Answer 3

If there is no wind resistance the best angle is 45 degrees. With wind resistance it's more like 40 degrees.

Answer 4

depends on the time it takes to get there. kinda the monkey in the tree example.

Answer 5

you know, you could have googled your homework and it probably would have been quicker... but it's 45 degrees