anonymous
  • anonymous
A projectile is thrown horizontallly from the top of a tower and strikes teh ground after 3 second at an angle of 45 degrees with the horizontal. Find the height of the tower and speed with which the body was projected. (Take "g" = 9.8 m/s^2)
Physics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
@Mani_Jha , @Ishaan94 please help.
anonymous
  • anonymous
@apoorvk , @srinidhijha, @AravindG please help
anonymous
  • anonymous
what is angle of projection? is the question written as it is?

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anonymous
  • anonymous
Yeah, the question is as it is as I have written here.
anonymous
  • anonymous
k
apoorvk
  • apoorvk
So when it strikes, basically the horizontal and vertical components of its velocity are equal (since it strikes the ground at 45 degrees) Now, let the initial velocity of projection be 'u' Now, the horizontal speed remains constant, but the vertical component increases from '0' to 'u', as it hits the ground. Now, vertical component (final) = v time of flight = sqrt(2gh) = 3 Find height, and then find 'u'.
anonymous
  • anonymous
How time of flight = sqrt(2gh)? Can u show the complete steps and other equations also?
apoorvk
  • apoorvk
LOL sorry - it would 'final' velocity, not time of flight - my bad lol.
anonymous
  • anonymous
Still i am not able to follow how is it equal to sqrt(2gh)? Can u show the workign of other steps also?
apoorvk
  • apoorvk
yeah, so use, (v-u)/g = t
anonymous
  • anonymous
hey is the speed 97 ? u know the answers?
anonymous
  • anonymous
Answer is 30 m/s
apoorvk
  • apoorvk
for a particle under free fall with initial velocity 'u=0'|dw:1338565657495:dw|
anonymous
  • anonymous
Got it. Can we not say that:\[T = 2u \cos \theta / g = \sqrt{2}u/g\]
apoorvk
  • apoorvk
yeah, v=30, and height = 45 metres, right?
apoorvk
  • apoorvk
well, that is for a ground-to-ground projectile @Aadarsh
anonymous
  • anonymous
@apoorvk bhai, ur answer is right.
anonymous
  • anonymous
@apoorvk bhai, please show all the steps.
anonymous
  • anonymous
look the pa
1 Attachment
anonymous
  • anonymous
Thanks apoorv bhai.
apoorvk
  • apoorvk
See, ye samajh mein aaya ki horizontal and vertical components at the time of hitting the ground same honge?
anonymous
  • anonymous
Yeh to samajh mein nahion aaya.....
anonymous
  • anonymous
samjha dijiye
anonymous
  • anonymous
see the attachment koi horizontal equal lene ki jarurat nh ihai
1 Attachment
anonymous
  • anonymous
samajh aaya?
anonymous
  • anonymous
isse pahle bhi ek attachment tha jisme U nikaala hai , dkho, Final velocity to 0 hi hoga naa jab wo gir jaaye to
anonymous
  • anonymous
Thakns srini bhaiiiiiiiii. samajh mein aa gaya.
anonymous
  • anonymous
k good
apoorvk
  • apoorvk
Sri, great calligraphy! :P
AravindG
  • AravindG
srry i am late :P
anonymous
  • anonymous
ha ha

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