Projectile Motion Range: Finding the Maximum Distance with Given Speed and Angle

In summary, to find the maximum range of a stone thrown upward at an angle of 31.0 degrees with a speed of 11.0m/s, you must first calculate the time it takes for the stone to reach its maximum height using the formula Vf=Vi+At and solving for time. Then, double this time and multiply it by the horizontal component, which is found by using the formula for cosine. This will give you the distance traveled by the stone over level ground. It is important to note that 45 degrees is the ideal angle for maximizing distance. Also, be sure to switch your calculator to degrees mode when solving this problem.
  • #1
Cole07
106
0
Question: How far will a stone travel over level ground if it is thrown upward at an angle of 31.0 degrees with respect to the horizontal and with a speed of 11.0m/s?


What is the maximum range that could be achieved with the same initial speed?


I have tried to solve this with no sucess what i have done is Drawn a triangle with a angle being 31.0 degrees the opposite side from the angle is Vy the adjacent side is Vx the hypothesis is 11.0m/s then i drew a chart with X and Y X side of the chart I have Vx= 10.06216594 a=0 then on the y side i have
Vy=-4.444414099 a=-9.8 the Vf=0m/s I used the formula Vf=Vi+At and solved for time and got
.453116428 then i used d=Vi*t+1/2At^2 and solved for D and got 2.132070497 for the first question (But was told this was not correct)

For the second question i doubled the time and multiplied by Vx D=10.06216594 (0.906232856) and got 9.118665377 but it was wrong !

Is there anyone who might be able to help me Please?
 
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  • #2
lmao, your calculator is in radians mode. Switch it to degrees mode :wink: After you have the time, don't forget it goes up and comes back down...

Your horizontal component will give you the distance, after you know the total time in the air.

45 degrees is a nice angle to maximize distance.
 
Last edited:
  • #3
Do you have to double time to solve for D with the D=ViT+1/2At^2 I know you must double it to get the maximum range (I am reworking the problem in degree mode I can't believe I did that!)
 
  • #4
Is the horzontial component 11.0cos(31.0) =9.428840308 this is the distance?
 

1. What is projectile motion range?

Projectile motion range is the horizontal distance traveled by an object in motion under the influence of gravity. It is the total distance covered by the object from the moment it is launched to the moment it hits the ground.

2. What factors affect the range of projectile motion?

The range of projectile motion is affected by the initial velocity, the angle of launch, and the acceleration due to gravity. The shape and weight of the object also play a role in determining the range.

3. How is the range of projectile motion calculated?

The range of projectile motion can be calculated using the equation R = (v^2 * sin2θ) / g, where R is the range, v is the initial velocity, θ is the launch angle, and g is the acceleration due to gravity.

4. Can the range of projectile motion be increased?

Yes, the range of projectile motion can be increased by increasing the initial velocity or changing the launch angle to a more optimal angle. Reducing air resistance can also increase the range of projectile motion.

5. What is the maximum range of projectile motion?

The maximum range of projectile motion occurs when the launch angle is 45 degrees. In this case, the range is equal to (v^2) / g, where v is the initial velocity and g is the acceleration due to gravity. Any launch angle less than 45 degrees will result in a shorter range, while angles greater than 45 degrees will also decrease the range.

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