Recent content by mustafamistik

I understand. Thanks for your help.

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

Yes, according to my calculations ##H_{max}## bigger than 3m

Sorry. You're right. ##0-(5*sqrt(5)*sin(45))^2=2*g*H_{max}## =4.46 Yes seems to be true .

##0-(5*sqrt(5))^2=2*H_{max}## ? Then ##62.5=H_{max}## But is result too big ?

I don't know how to do that.

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.

Thanks. I don't know how to discuss part c, what logic should I think?

Understood. Actually i wonder if my solution way is correct. For part a.

I rounded it. It comes from 4,9/5*tan^2... equation.

It's attached.

Is my solving way correct for part a ? I don't know how to discuss part c, what logic should I think?

I stuck on part c and d. My attemption is attached.

Thank you and @haruspex for helping solve and understand this question. Also thanks for your interest.

Is that equation true ?