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Can you show how you arrived at your solution attempt? What does "costheta(t)" represent?Kingyou123 said:Homework Statement
Uploaded
Homework Equations
x(t)=xo+v0xT+1/2axt^2
The Attempt at a Solution
(2Vo)costheta(t)=R, How would I get rid of T?
Velocity Intial of x is equal to vocostheta, should I be using range =(vo^2sin2(theta))/gravity. I'm really lost right now...gneill said:Can you show how you arrived at your solution attempt? What does "costheta(t)" represent?
Okay, that's a bit better. The Range equation is a good approach. You should place the entire argument of a function within the parentheses to make it clear what the function argument is. Thus:Kingyou123 said:Velocity Intial of x is equal to vocostheta, should I be using range =(vo^2sin2(theta))/gravity. I'm really lost right now...
Would it be 2R since the velocity is doubled?gneill said:Okay, that's a bit better. The Range equation is a good approach. You should place the entire argument of a function within the parentheses to make it clear what the function argument is. Thus:
R = (vo2/g) sin(2θ)
(Note that you can use the ##x_2## and ##x^2## icons in the edit panel header to invoke superscripts and subscripts, and greek letters and other symbols can be selected from the ##\Sigma## icon's menu)
The range equation gives you the range of a projectile that's launched with a given velocity ##v_o## at a given angle ##\theta##. So what happens to the range if you double the launch velocity?
For the second part of the question, take a browse though your table of trig identities then ponder what happens to sin and cos if the angle is adjusted as specified in the question.
Can you justify that with an argument based upon the range equation? I won't confirm or deny a guess...Kingyou123 said:Would it be 2R since the velocity is doubled?
It is not correct. Does the range equation use ##v_o## or ##v_o^2##? How does squaring a doubled value affect the net result?Kingyou123 said:I plugged 10 for the in initial velocity so doubling that would make 20/g therefore the outcome would be twice as great. Right logic or I'm I completely off/
ohhhhhhh would it be 4R cause by doubling 1 you get 2 and 2^2 is 4 so in this case quadruplicating the R.gneill said:It is not correct. Does the range equation use ##v_o## or ##v_o^2##? How does squaring a doubled value affect the net result?
Bingo! It always pays to consider the equation involved and not rely on instinct aloneKingyou123 said:ohhhhhhh would it be 4R cause by doubling 1 you get 2 and 2^2 is 4 so in this case quadruplicating the R.
Thank you so much it makes so much more sense now :) Would you be up to help with the rest of homework when I get stuck?gneill said:Bingo! It always pays to consider the equation involved and not rely on instinct alone
For the second part would the angle be greater causing the distance to decrease? If I plug 30 in for theta, sin(90-30), the angle is greater than the previous 30, but if I plug 85 for theta the answer would less than...Kingyou123 said:Thank you so much it makes so much more sense now :) Would you be up to help with the rest of homework when I get stuck?
The formula for calculating the range of a projectile is R = (v^2*sin(2θ))/g, where R is the range, v is the initial velocity of the projectile, θ is the angle of launch, and g is the acceleration due to gravity.
The maximum range of a projectile can be calculated by setting the angle of launch, θ, to 45 degrees. The formula then simplifies to R = v^2/g, where v is the initial velocity of the projectile and g is the acceleration due to gravity.
The key factors that affect the range of a projectile are the initial velocity, angle of launch, and the acceleration due to gravity. Air resistance can also impact the range, but it is often negligible in projectile motion calculations.
No, the range of a projectile cannot be negative. It is a measure of the horizontal distance traveled by the projectile and therefore must always be a positive value.
Increasing the initial velocity of a projectile will result in a longer range. This is because the range formula is directly proportional to the square of the initial velocity, meaning that as the initial velocity increases, the range will also increase.