y=1/x & dy/dx=-1/x^2. So slope of the normal at the pt. of reboumd is = x^2 = 4 or tan(α)=4 or cos(α)=1/√17. Velocity along normal=(0.97014)√(2g*144.5). Eqn of parabolic path with the point of rebound as the origin: y = 4x - x^2/32 after substituting all relevant values. At x=64 the peak is reached with h = 128+0.5 = 128.5 above the original x-axis. Unfortunately, this option isn't there. So pl. tell me where I'm going wrong. Ajit

Launch angle is the angle between the initial velocity direction and the x-axis i.e. the direction of the normal & the x-axis in our case or tan(α)=4 so the launch angle is ~ 75.964 deg. to the horizontal and the general equation is: y = xtan(α)-(g/2)(x/ucos(α))^2 where the point of rebound is the origin.

sorry i think i made amistake. let me try again. tangent to 1/x at x=2 has a slope of m1=-1/4. slope of the normal=m2. m1*m2=-1 => m2=4. => slope of the normal is 4. let a be the launch angle. let b be the angle between normal and the y-axis. tan(b)=1/4. we also know: a+2b=90 => 2b=90-a => tan(2b)=tan(90-a) => 2tan(b)/(1-tan^2(b))=cot(a) => 2tan(b)/(1-tan^2(b))=cot(a) => 2(1/4)/(1-1/16)=cot(a) => 1/2 /(15/16)=cot(a) => 8/15 =cot(a) it is a 8-15-17 triangle!!! => tan(a)= 15/8 => sin(a) = 15/17 => cos(a) = 8/17 so the launch angle has a cos of 8/17. rest should follow. -binary

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y=1/x & dy/dx=-1/x^2. So slope of the normal at the pt. of reboumd is = x^2 = 4 or tan(α)=4 or cos(α)=1/√17. Velocity along normal=(0.97014)√(2g*144.5). Eqn of parabolic path with the point of rebound as the origin: y = 4x - x^2/32 after substituting all relevant values. At x=64 the peak is reached with h = 128+0.5 = 128.5 above the original x-axis. Unfortunately, this option isn't there. So pl. tell me where I'm going wrong.

ReplyDeleteAjit

"So slope of the normal at the pt. of reboumd is = x^2 = 4"

ReplyDeletethen the laungh angle should be (90 - 2*alpha). is that not the case?

launch angle: angle btw the rebound and the horizontal.

-binary

Launch angle is the angle between the initial velocity direction and the x-axis i.e. the direction of the normal & the x-axis in our case or tan(α)=4 so the launch angle is ~ 75.964 deg. to the horizontal and the general equation is:

ReplyDeletey = xtan(α)-(g/2)(x/ucos(α))^2 where the point of rebound is the origin.

i still think "tan(α)=4" is the slope of the normal => launch angle is 90 - 2α.

ReplyDelete-binary

sorry i think i made amistake. let me try again. tangent to 1/x at x=2 has a slope of m1=-1/4. slope of the normal=m2. m1*m2=-1 => m2=4. => slope of the normal is 4. let a be the launch angle. let b be the angle between normal and the y-axis. tan(b)=1/4. we also know:

ReplyDeletea+2b=90 => 2b=90-a

=> tan(2b)=tan(90-a)

=> 2tan(b)/(1-tan^2(b))=cot(a)

=> 2tan(b)/(1-tan^2(b))=cot(a)

=> 2(1/4)/(1-1/16)=cot(a)

=> 1/2 /(15/16)=cot(a)

=> 8/15 =cot(a) it is a 8-15-17 triangle!!!

=> tan(a)= 15/8

=> sin(a) = 15/17

=> cos(a) = 8/17

so the launch angle has a cos of 8/17. rest should follow.

-binary

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ReplyDelete