What Is the Minimum Launch Angle to Hit a Target 200ft Away?

[AirForceOne]

Suppose we have a cannon at a height of 3 ft and want the projectile to hit the ground a distance d=200ft downfield. The launch speed v0 is 100ft/s. What is the minimum needed launch angle and the corresponding total time of flight?

I wasn't able to figure this problem out, so I looked at the answer. The answer is 19.06 degrees and 2.116 seconds. Using these values in the kinematic equations, I calculated that at x=200 ft, the height of the projectile is y=50ft. When x=200ft, shouldn't y=0ft? I thought the problem is asking to find the "smallest angle needed to hit the ground 200 ft downfield". I don't understand the meaning of minimum/maximum. To me, the projectile either hits (200,0) or it doesn't. If the problem gives us the initial velocity, shouldn't there only be one value of the angle to hit (200,0) instead of multiple like the words "minimum/maximum angle" is suggesting?

EDIT: I just realized that the problem could be asking for the smallest angle for the projectile to reach the distance of 200ft.That is, it's okay if the projectile overshoots the 200ft. However, wouldn't that mean finding the angle to produce the max distance? So the angle would be 45? But that's different from the correct answer of 19.06...I'm going crazy.

 

[tiny-tim]

Hi AirForceOne!

… I don't understand the meaning of minimum/maximum. To me, the projectile either hits (200,0) or it doesn't. If the problem gives us the initial velocity, shouldn't there only be one value of the angle to hit (200,0) instead of multiple like the words "minimum/maximum angle" is suggesting?

You should get a quadratic equation for the angle, with two (not "multiple") solutions …

a "smash" (minimum angle) and a "lob" (maximum angle). ​

 

[xts]

Just few comments:

As Tiny-tim noticed, it is more tennis game example, than cannon one (as an opposition to many textbook excercises about tennis balls traveling at 800m/s)

I am getting sick if my students bring me answers "19.06 degrees" in such cases. You are not able to measure the angle of tennis ball with such precision. 19° would be a bit exaggerated, 20° seems realistic rounding.

Never say something like "19.06 degrees". If you like to use traditional - Babylonian - measure (degrees rather than radians) then stay with traditional units and say: 19°1'
"19.06 degrees" is as inappropriate as you saying you are "6.16 feet tall" rather than "six feet two inches"

 

[Rayquesto]

I suppose you could use dy=Vo,y + 1/2gt^2 if the cannon is 3 meters high, then dy=-3meters, when the launch hits the ground, but I know there's an easier way.

Galileo, I think, also found that dx=V0^2 times sin(2theta)/g

dx will be 200 meters at some theta angle

200=(100m/s)^2 times sin(2theta)/g

solve for theta.

 

[Rayquesto]

eh sorry. I just noticed that this doesn't account for the 3 meter height, but maybe you can use that equation somehow. try it. again, IM REALLY SORRY for giving you a false answer.

 

[Rayquesto]

for fall off cliff models, use sqrt(2y/g) to find time from max height to when the ball goes to the ground which means that you could say that Vy=V0sin(theta) - gt where 0m/s=Vy at max height and that

(time to reach the max height= V0sin(theta)/g) where

dy,max=v0,y(v0,y/g) - 1/2g(v0,y/g)^2

and time to reach max height= V0sin(theta)/g time to go from max height to the ground= sqrt(2(dy,max)/g)

knowing this can be essential to finding the angle if you know range. if you know how long this must be in the air, don't you think that any equation that describes range will determine 200 meters.

so, 200meters= V0cos(theta)(time to reach max height) + V0(time to hit ground)
then plug in the times and you should be able to solve fore theta. the second v0 in the above equation may not be right. What do you think is the velocity equation as it goes down in the x direction? god this problem is crazy.

 

[AirForceOne]

Hi AirForceOne!


You should get a quadratic equation for the angle, with two (not "multiple") solutions …

a "smash" (minimum angle) and a "lob" (maximum angle). ​

What's the equation...? I just have cos(theta) = v0x/v0 or sin(theta) = v0y/v0.

Thanks.

 

[Rayquesto]

oh wait, for the second V0 make it V0costheta then solve for theta since it is the same velocity coming down than it is up.

 

[tiny-tim]

Hi AirForceOne!

(just got up :zzz: …)

What's the equation...? I just have cos(theta) = v0x/v0 or sin(theta) = v0y/v0.

You need two equations, one for the x direction and one for the y direction, both involving t.

Then you solve them as simultaneous equations (because they use the same t).

Your equations will be the standard https://www.physicsforums.com/library.php?do=view_item&itemid=204" equations, with a = 0 for the x direction, and a = -g for the y direction.

Show us what you get. ​