math, week-4, act-4, q-1.

[Geetaanjali GNS]

1 point

A stone is thrown with an initial speed uu (m/s) as shown in figure 3. The height of the stone’s trajectory above the ground is, H(t)=−5t2+12utH(t)=−5t2+12ut (here tt is the time of flight). If the highest point in air that the stone can reach is 5m above the ground, then calculate the initial speed uu.

 20 m/s

 10 m/s

 12 m/s

 2 m/s

i am getting answer as 52.

can someone tell me how to solve this?

[Vedansh Institute]

Answer is 20

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[Geetaanjali GNS]

yeah, I know what the answer is, but could someone tell me the procedure for solving this? I want to know the correct way of solving it.

[Charles George]

https://groups.google.com/a/nptel.iitm.ac.in/g/ma1001-discuss/c/F2MgCsdyHyo

[Malabika Guha Mustafi]

For solving the problem , it is better approach to replace t ( max)  with u/20. [ you need to compute u not t]

Then put 5 in place of H(t)[which is also max value means vertex].

Then you will get a equation of u only.

compute u directly.

[Sameeksha Kothari]

How did u get u/20?

[Malabika Guha Mustafi]

what is the formula? Is it a open up or open down parabola?

[Anand Iyer]

-b/2a is the x-coordinate of the highest point of the parabola.  Substitute it in the equation to get the maximum height.  Equate this to 5 (given).  

From this new equation, you'll be able to find u.

[Debajyoti Biswas]

Thanks Malabika and Anand, as all of us understand, it is a downward opening parabola and x-axis is time t, so we have at what time it reaches the vertex or the highest point, and we know the max. height, from there you can find u