Week 4 . Activity 4.4 . Question no.1

[Aseem Anand]

Hi,

Q)A stone is thrown with an initial speed u (m/s) as shown in figure 3. The height of the stone’s trajectory above the ground is, H(t)=−5t^2+1/2utH(t) (here t 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 u.  

can you please help me ith the solution of this question?

Regards,

Aseem Anand

[Malabika Guha Mustafi]

https://groups.google.com/a/nptel.iitm.ac.in/g/ma1001-discuss/c/C4aHK1uPmUU/m/j2KjK_8gAAAJ

[Kennedyraj 05]

Even after that I couldn't understand can you please share the solution.

[Malabika Guha Mustafi]

@kennedyraj, first compute the x coordinate of vertex. share the result.Then I'll help you

[Kennedyraj 05]

I have computed that the x coordinate of vertex as u/20 using the formula (-b/2a). Please do guide forth.

[Malabika Guha Mustafi]

Good. Then see , the y coordinate value =5( since it is maximum at vertex),. you computed x coordinate value at vertex which is t.

Now, put both the value in equation and compute u. Hope, it helps you.

[Kennedyraj 05]

@Malabika Guha Mustafi, Thank you for helping me to compute the answer.

[Malabika Guha Mustafi]

Welcome

[shyam sundar]

Hello Malabika ji,

Thank you! I want to understand how to got x axis as t and y axis as H(t). I took x axis as t and y axis as H. Hence I was getting an equation in H and u.

[SUHANA PARVEEN S]

That's why our eqn is ax^2+bx+c... 

[Jaya Krishna Vamsi Krishnamsetti]