range for 45 degree launch angle

[Addallat]

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Top of page 30 of Berkley Review: "Maximum range is 2H possible only 2ith a 45 degree launch angle"

then in example 1.14
page 31
"The range for a 45 degree launch is four times the maximum height"


is it 2x the maximum height or 4x the maximum height?

What am I missing here?

 

[GolfUSA]

4x maximum height. It's a decent thing to memorize. I have a notecard for 30 degrees (6.7 R/H), 45 degrees (4.0 R/H), and 60 degrees (2.4 R/H) which is stated in the TBR physics chapter.

 

[tdod]

you can also think about this logically; the average horizontal speed is constant 2 the average horizontal speed. Furthermore, horizontal velocity is constant, whereas vertical velocity points upwards for only the first half of the launch.

 

[Addallat]

Top of page 30 of Berkley Review: "Maximum range is 2H possible only 2ith a 45 degree launch angle"

What am I missing here?

 

[BerkReviewTeach]

Top of page 30 of Berkley Review: "Maximum range is 2H possible only 2ith a 45 degree launch angle"

What am I missing here?

The top of the page for me reads "The maximum height that a projectile can possibly reach is h (possible only with a launch angle of 90 degrees), while the maximum range is 2h (possible only when the launch angle is 45 degrees). It then goes on to say that the height when launched at 45 degrees is h/2. These are statements referring to Figure 1-20, which shows the different ranges when launching at the same speed but from different angles.

What this all means is that hmax at 45 degrees is h/2 (where h refers to the height that would have been reached with a 90-degree launch) and the range is 2h. The ratio of the range (R) to the maximum height (hmax) for a 45-degree launch is 2h : h/2 which is equal to 4h : h = 4 : 1. That correlates with everything else.

So if the launch angle was 45 degrees with a speed of 27 m/s for instance, we know that vx is about 20 m/s so ascent time would be 2s (from the v = 10t shortcut) and thus hapex would be 20 m (from 5t^2) which at last means R = 80 (4 x 20). These shortcuts allow you to find all sorts of quick answers.

 

[el_duderino]

What the previous poster said. But more simply:

When shot straight up, maximum height is h. When shot at 45 degrees, maximum height is h/2. Four times h/2 is 2h.

 

[NextStepTutor_1]

Use the kinematic equations to help you out as well! I we know that the angle is 45, we know the horizontal and vertical velocities are the same as well--let's call them v.

So, for the vertical motion at the top of the trajectory:
a = (vf - vi) / t

-10 = (0 - v)/t

t = v/10
so total flight time is v/5.

So, for horizontal motion, range is v*t = v^2 / 5.

The vertical distance is :

d = 1/2*a*t^2 + v*t

d = -5*(v/10)^2 + v * (v/10)

d = -v^2 /20 + v^2 / 10

d = v^2/20.

Thus, if we compare the two displacements, we get that the horizontal range is 4x the vertical range.