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Is my solving way correct for part a ? I don't know how to discuss part c, what logic should I think?PeroK said:Part c) says "discuss" and I don't see any discussion. I can't work out what your answer to d) is either.
What is your answer for part a)?mustafamistik said:Is my solving way correct for part a ? I don't know how to discuss part c, what logic should I think?
It's attached.PeroK said:What is your answer for part a)?
45 degrees?mustafamistik said:It's attached.
I rounded it. It comes from 4,9/5*tan^2... equation.PeroK said:45 degrees?
That said, this is a strange problem. The answer must be close to 45 degrees, but it cannot be exactly 45 degrees, unless you take ##g = 10 m/s^2##.
Rounding or not makes a big difference to part c).mustafamistik said:I rounded it. It comes from 4,9/5*tan^2... equation.
Understood. Actually i wonder if my solution way is correct. For part a.PeroK said:Rounding or not makes a big difference to part c).
It looks okay to me.mustafamistik said:Understood. Actually i wonder if my solution way is correct. For part a.
Thanks. I don't know how to discuss part c, what logic should I think?PeroK said:It looks okay to me.
It wants calculations: angle, speed gives max height.mustafamistik said:Thanks. I don't know how to discuss part c, what logic should I think?
Can you not tidy that up? You should get a simple formula for max height if ##\theta = 45°##mustafamistik said:In maximum height vertical speed must equal to 0.
##V_0*sin(45)/g= t##
##(V_0*sin(45)*t)-(1/2*g*t^2)=H_{max}##
##(V_0*sin(45)*(V_0*sin(45)/g))-(1/2*g*(V_0*sin(45)/g)^2)=H_{max}## ...
Is this way true ? for part C.
I don't know how to do that.PeroK said:Can you not tidy that up? You should get a simple formula for max height if ##\theta = 45°##
It's basic kinematics. ##v^2 - u^2 = 2as##mustafamistik said:I don't know how to do that.
##0-(5*sqrt(5))^2=2*H_{max}## ?PeroK said:It's basic kinematics. ##v^2 - u^2 = 2as##
Or, in this case ##v_y^2 - u_y^2 = 2a_ys_y##
Isn't there a little thing called gravity involved in the calculation? And isn't the angle involved too?mustafamistik said:##0-(5*sqrt(5))^2=2*H_{max}## ?
Then ##62.5=H_{max}##
But is result too big ?
Sorry. You're right.PeroK said:Isn't there a little thing called gravity involved in the calculation? And isn't the angle involved too?
I thought you were trying to find whether ##H_{max} = 3m## or not.mustafamistik said:Sorry. You're right.
##0-(5*sqrt(5)*sin(45))^2=2*g*H_{max}## =4.46
Yes seems to be true .
Yes, according to my calculations ##H_{max}## bigger than 3mPeroK said:I thought you were trying to find whether ##H_{max} = 3m## or not.
Okay, but you may want to check your calculations: ##4.46m## is too high.mustafamistik said:Yes, according to my calculations ##H_{max}## bigger than 3m
I assume that's right. It's an odd question, as it has little relevance to the specific problem.mustafamistik said:For part D. Radial component is zero, tangential component is ##g## of acceleration. Am i wrong?
I understand. Thanks for your help.PeroK said:I assume that's right. It's an odd question, as it has little relevance to the specific problem.
Throwing an object in one dimension refers to throwing it in a straight line, while throwing an object in two dimensions refers to throwing it in a curved path, taking into account both horizontal and vertical components of motion.
Air resistance can affect the motion of a thrown object in two dimensions by slowing it down and altering its trajectory. This is because air resistance acts in the opposite direction of the object's motion, causing it to lose speed and deviate from its intended path.
The horizontal component of a thrown object's velocity can be calculated using the equation Vx = Vcosθ, where V is the initial velocity and θ is the angle of the throw. The vertical component can be calculated using Vy = Vsinθ.
The angle of the throw can greatly affect the range of a thrown object in two dimensions. The optimal angle for maximum range is 45 degrees, as this creates the perfect balance between the horizontal and vertical components of velocity. Throwing at a higher or lower angle will result in a shorter range.
Projectile motion refers to the motion of an object that is thrown or launched into the air, while parabolic motion specifically refers to the curved path that the object follows in two dimensions. All objects in projectile motion follow a parabolic path, but not all objects in parabolic motion are necessarily projectiles.