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)

Question

anonymous · Answer

@Mani_Jha , @Ishaan94 please help.

anonymous · Answer

@apoorvk , @srinidhijha, @AravindG  please help

anonymous · Answer

what is angle of projection? is the question written as it is?

anonymous · Answer

Yeah, the question is as it is as I have written here.

anonymous · Answer

k

apoorvk · Answer

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 · Answer

How time of flight = sqrt(2gh)? Can u show the complete steps and other equations also?

apoorvk · Answer

LOL sorry - it would 'final' velocity, not time of flight - my bad lol.

anonymous · Answer

Still i am not able to follow how is it equal to sqrt(2gh)? Can u show the workign of other steps also?

apoorvk · Answer

yeah, so use, (v-u)/g = t

anonymous · Answer

hey is the speed 97 ? u know the answers?

anonymous · Answer

Answer is 30 m/s

apoorvk · Answer

for a particle under free fall with initial velocity 'u=0'|dw:1338565657495:dw|