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 2 years ago
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)
 2 years ago
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)

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Aadarsh
 2 years ago
Best ResponseYou've already chosen the best response.0@Mani_Jha , @Ishaan94 please help.

Aadarsh
 2 years ago
Best ResponseYou've already chosen the best response.0@apoorvk , @srinidhijha, @AravindG please help

srinidhijha
 2 years ago
Best ResponseYou've already chosen the best response.1what is angle of projection? is the question written as it is?

Aadarsh
 2 years ago
Best ResponseYou've already chosen the best response.0Yeah, the question is as it is as I have written here.

apoorvk
 2 years ago
Best ResponseYou've already chosen the best response.1So 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'.

Aadarsh
 2 years ago
Best ResponseYou've already chosen the best response.0How time of flight = sqrt(2gh)? Can u show the complete steps and other equations also?

apoorvk
 2 years ago
Best ResponseYou've already chosen the best response.1LOL sorry  it would 'final' velocity, not time of flight  my bad lol.

Aadarsh
 2 years ago
Best ResponseYou've already chosen the best response.0Still i am not able to follow how is it equal to sqrt(2gh)? Can u show the workign of other steps also?

apoorvk
 2 years ago
Best ResponseYou've already chosen the best response.1yeah, so use, (vu)/g = t

srinidhijha
 2 years ago
Best ResponseYou've already chosen the best response.1hey is the speed 97 ? u know the answers?

apoorvk
 2 years ago
Best ResponseYou've already chosen the best response.1for a particle under free fall with initial velocity 'u=0'dw:1338565657495:dw

Aadarsh
 2 years ago
Best ResponseYou've already chosen the best response.0Got it. Can we not say that:\[T = 2u \cos \theta / g = \sqrt{2}u/g\]

apoorvk
 2 years ago
Best ResponseYou've already chosen the best response.1yeah, v=30, and height = 45 metres, right?

apoorvk
 2 years ago
Best ResponseYou've already chosen the best response.1well, that is for a groundtoground projectile @Aadarsh

Aadarsh
 2 years ago
Best ResponseYou've already chosen the best response.0@apoorvk bhai, ur answer is right.

Aadarsh
 2 years ago
Best ResponseYou've already chosen the best response.0@apoorvk bhai, please show all the steps.

apoorvk
 2 years ago
Best ResponseYou've already chosen the best response.1See, ye samajh mein aaya ki horizontal and vertical components at the time of hitting the ground same honge?

Aadarsh
 2 years ago
Best ResponseYou've already chosen the best response.0Yeh to samajh mein nahion aaya.....

srinidhijha
 2 years ago
Best ResponseYou've already chosen the best response.1see the attachment koi horizontal equal lene ki jarurat nh ihai

srinidhijha
 2 years ago
Best ResponseYou've already chosen the best response.1isse pahle bhi ek attachment tha jisme U nikaala hai , dkho, Final velocity to 0 hi hoga naa jab wo gir jaaye to

Aadarsh
 2 years ago
Best ResponseYou've already chosen the best response.0Thakns srini bhaiiiiiiiii. samajh mein aa gaya.

apoorvk
 2 years ago
Best ResponseYou've already chosen the best response.1Sri, great calligraphy! :P
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