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- Thread starter mustafamistik
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In summary, the problem asks for an equation for the maximum height of a Vertical Slider moving at a speed of 10 m/s.

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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:

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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?

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It's attached.PeroK said:What is your answer for part a)?

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45 degrees?mustafamistik said:It's attached.

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##.

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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##.

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Rounding or not makes a big difference to part c).mustafamistik said:I rounded it. It comes from 4,9/5*tan^2... equation.

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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).

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It looks okay to me.mustafamistik said:Understood. Actually i wonder if my solution way is correct. For part a.

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Thanks. I don't know how to discuss part c, what logic should I think?PeroK said:It looks okay to me.

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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?

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##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.

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Can you not tidy that up? You should get a simple formula for max height if ##\theta = 45°##mustafamistik said:

##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.

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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°##

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It's basic kinematics. ##v^2 - u^2 = 2as##mustafamistik said:I don't know how to do that.

Or, in this case ##v_y^2 - u_y^2 = 2a_ys_y##

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##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##

Then ##62.5=H_{max}##

But is result too big ?

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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 ?

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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?

##0-(5*sqrt(5)*sin(45))^2=2*g*H_{max}## =4.46

Yes seems to be true .

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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 .

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Yes, according to my calculations ##H_{max}## bigger than 3mPeroK said:I thought you were trying to find whether ##H_{max} = 3m## or not.

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For part D. Radial component is zero, tangential component is ##g## of acceleration. Am i wrong?

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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

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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?

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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.

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