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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
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@Mani_Jha , @Ishaan94 please help.
what is angle of projection? is the question written as it is?

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Other answers:

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

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