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A body projected vertically up crosses points A & B separated by 28 m with velocities one third and one fourth of initial velocity respectively. What is the maximum height reached by it above the ground.

Please answer it fast sir

Thank You

Hello Ankit,

Use the kinematic formula

Vf^2 = Vi^2 + 2*a*y

We know that Vfa = Vi/3 and Vfb = Vi/4.

At point A:

Vfa^2 = Vi^2 + 2*(-9.8 m/s^2)*Ya

Solving for Vi^2

Vi^2 = Vfa^2 + 2*(9.8 m/s^2)*Ya

Substituting for Vfa

Vi^2 = (Vi/3)^2 + 2*(9.8 m/s^2)*Ya

Vi^2 = (1/9)*Vi^2 + 19.6 m/s^2*Ya

(8/9)*Vi^2 = 19.6 m/s^2*Ya

Vi^2 = (9/8)*19.6 m/s^2*Ya = 22.05 m/s^2*Ya

At point B:

Vfb^2 = Vi^2 + 2*(-9.8 m/s^2)*(Ya+28 m)

Solving for Vi^2

Vi^2 = Vfb^2 + 2*(9.8 m/s^2)*(Ya+28 m)

Substituting for Vfb

Vi^2 = (Vi/4)^2 + 2*(9.8 m/s^2)*(Ya+28 m)

Vi^2 = (1/16)*Vi^2 + 2*(9.8 m/s^2)*(Ya+28 m)

(15/16)*Vi^2 = 2*(9.8 m/s^2)*(Ya+28 m)

I'm going to leave some work for you to do in return for quickly answering. You have one more step to do to the above expression to give you an expression for Vi^2 equivalent to the bottom line of the point A analysis. Then you will have 2 expressions for Vi^2. Set those equal to each other and work out what Ya is. Knowing Ya, you can then plug it into the bottom line of the point A analysis to find Vi. Knowing Vi, you can plug it into the basic kinematic formula (the top equation above), with Vf=0, to find max height.

I hope this helps,

Steve

Physics

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