Aadarsh
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)
Delete
Share
This Question is Open
Aadarsh
Best Response
You've already chosen the best response.
0
@Mani_Jha , @Ishaan94 please help.
Aadarsh
Best Response
You've already chosen the best response.
0
@apoorvk , @srinidhijha, @AravindG please help
srinidhijha
Best Response
You've already chosen the best response.
1
what is angle of projection? is the question written as it is?
Aadarsh
Best Response
You've already chosen the best response.
0
Yeah, the question is as it is as I have written here.
srinidhijha
Best Response
You've already chosen the best response.
1
k
apoorvk
Best Response
You'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'.
Aadarsh
Best Response
You've already chosen the best response.
0
How time of flight = sqrt(2gh)? Can u show the complete steps and other equations also?
apoorvk
Best Response
You've already chosen the best response.
1
LOL sorry - it would 'final' velocity, not time of flight - my bad lol.
Aadarsh
Best Response
You'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?
apoorvk
Best Response
You've already chosen the best response.
1
yeah, so use, (v-u)/g = t
srinidhijha
Best Response
You've already chosen the best response.
1
hey is the speed 97 ? u know the answers?
Aadarsh
Best Response
You've already chosen the best response.
0
Answer is 30 m/s
apoorvk
Best Response
You've already chosen the best response.
1
for a particle under free fall with initial velocity 'u=0'|dw:1338565657495:dw|
Aadarsh
Best Response
You've already chosen the best response.
0
Got it. Can we not say that:\[T = 2u \cos \theta / g = \sqrt{2}u/g\]
apoorvk
Best Response
You've already chosen the best response.
1
yeah, v=30, and height = 45 metres, right?
apoorvk
Best Response
You've already chosen the best response.
1
well, that is for a ground-to-ground projectile @Aadarsh
Aadarsh
Best Response
You've already chosen the best response.
0
@apoorvk bhai, ur answer is right.
Aadarsh
Best Response
You've already chosen the best response.
0
@apoorvk bhai, please show all the steps.
srinidhijha
Best Response
You've already chosen the best response.
1
look the pa
Aadarsh
Best Response
You've already chosen the best response.
0
Thanks apoorv bhai.
apoorvk
Best Response
You'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?
Aadarsh
Best Response
You've already chosen the best response.
0
Yeh to samajh mein nahion aaya.....
Aadarsh
Best Response
You've already chosen the best response.
0
samjha dijiye
srinidhijha
Best Response
You've already chosen the best response.
1
see the attachment koi horizontal equal lene ki jarurat nh ihai
srinidhijha
Best Response
You've already chosen the best response.
1
samajh aaya?
srinidhijha
Best Response
You'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
Aadarsh
Best Response
You've already chosen the best response.
0
Thakns srini bhaiiiiiiiii. samajh mein aa gaya.
srinidhijha
Best Response
You've already chosen the best response.
1
k good
apoorvk
Best Response
You've already chosen the best response.
1
Sri, great calligraphy! :P
AravindG
Best Response
You've already chosen the best response.
0
srry i am late :P
srinidhijha
Best Response
You've already chosen the best response.
1
ha ha