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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)
 one year ago
 one year 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)
 one year ago
 one year ago

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AadarshBest ResponseYou've already chosen the best response.0
@Mani_Jha , @Ishaan94 please help.
 one year ago

AadarshBest ResponseYou've already chosen the best response.0
@apoorvk , @srinidhijha, @AravindG please help
 one year ago

srinidhijhaBest ResponseYou've already chosen the best response.1
what is angle of projection? is the question written as it is?
 one year ago

AadarshBest ResponseYou've already chosen the best response.0
Yeah, the question is as it is as I have written here.
 one year ago

apoorvkBest ResponseYou've already chosen the best response.1
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'.
 one year ago

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

apoorvkBest ResponseYou've already chosen the best response.1
LOL sorry  it would 'final' velocity, not time of flight  my bad lol.
 one year ago

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

apoorvkBest ResponseYou've already chosen the best response.1
yeah, so use, (vu)/g = t
 one year ago

srinidhijhaBest ResponseYou've already chosen the best response.1
hey is the speed 97 ? u know the answers?
 one year ago

apoorvkBest ResponseYou've already chosen the best response.1
for a particle under free fall with initial velocity 'u=0'dw:1338565657495:dw
 one year ago

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

apoorvkBest ResponseYou've already chosen the best response.1
yeah, v=30, and height = 45 metres, right?
 one year ago

apoorvkBest ResponseYou've already chosen the best response.1
well, that is for a groundtoground projectile @Aadarsh
 one year ago

AadarshBest ResponseYou've already chosen the best response.0
@apoorvk bhai, ur answer is right.
 one year ago

AadarshBest ResponseYou've already chosen the best response.0
@apoorvk bhai, please show all the steps.
 one year ago

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

AadarshBest ResponseYou've already chosen the best response.0
Yeh to samajh mein nahion aaya.....
 one year ago

srinidhijhaBest ResponseYou've already chosen the best response.1
see the attachment koi horizontal equal lene ki jarurat nh ihai
 one year ago

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

AadarshBest ResponseYou've already chosen the best response.0
Thakns srini bhaiiiiiiiii. samajh mein aa gaya.
 one year ago

apoorvkBest ResponseYou've already chosen the best response.1
Sri, great calligraphy! :P
 one year ago
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