Help With finding initial vertical speed

  • #1
mithilsheth
22
0

Homework Statement



A Ball is thrown upwards. What is its initial vertical speed? The acceleration of gravity is 9.8 m/s2 and maximum height is 7.2 m ,. Neglect air resistance.


Homework Equations



Vfinal squared = velocity initial squared + 2g times change in y

The Attempt at a Solution



velocity final squared = velocity initial squared + 2 (-9.8)(7.2)
VF^2= VI^2 + -141.12
what now?/
 

Answers and Replies

  • #2
Doc Al
Mentor
45,410
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What's the final velocity? (When it reaches the max height.)
 
  • #3
mithilsheth
22
0
the final velocity is not given. all that is given is v0, max height is 7.2 meters, and that the acceleration of gravity is 9.8 m/s/s/.
 
  • #4
mithilsheth
22
0
is the finaly velocity 0?
 
  • #6
mithilsheth
22
0
ohh ok... so would i use the same formula i used above to find the initial vertical speed?
 
  • #7
mithilsheth
22
0
what equation should i use to calculate the initial vertical speed?
 
  • #8
Doc Al
Mentor
45,410
1,841
ohh ok... so would i use the same formula i used above to find the initial vertical speed?
Just use that equation you started with and solve for the initial speed.
 
  • #9
mithilsheth
22
0
how do i set it to be that velocity initial squared is on the left side of the equal sign, and velocity final squared is on on the right side?
 
  • #10
Doc Al
Mentor
45,410
1,841
how do i set it to be that velocity initial squared is on the left side of the equal sign, and velocity final squared is on on the right side?
The final velocity is zero, so your equation becomes:

0 = Vi2 - 141.12

So how to get Vi2 alone? How can you get rid of that - 141.12?
 
  • #11
mithilsheth
22
0
you add 141.12 to the 0 and then find square root of 141.12?
 
  • #12
Doc Al
Mentor
45,410
1,841
you add 141.12 to the 0 and then find square root of 141.12?
Sure! Basic algebra.
 
  • #13
mithilsheth
22
0
so the initial vertical speed would be 11.8793939239 m/s ?
 
  • #14
Doc Al
Mentor
45,410
1,841
so the initial vertical speed would be 11.8793939239 m/s ?
Yes. But please round off to a reasonable number of significant figures. (2 or 3 is plenty.)
 
  • #15
mithilsheth
22
0
ok, i have one more question i need help with....
 

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