Question

A ball is dropped from the top of the tower falls first half height of tower in 10 seconds

Answer 1

Let the time during which the ball remains in air be T seconds

Then we have h=1/2 gT^2 …(1)

Now distance travelled by ball in nth second is given by Sn=g/2(2n-1)

h/2=g/2(2T-1)…(2)

Substituting h from (1) in (2) we get

T^2=4T-2

T^2–4T+2=0

Which gives T=3.41 seconds

312 Views · View Upvoters

Upvote · 2

Share

 

Comment...



Amritpritam Kar, studied at Dav Public School,pkt,bbsr

Answered May 26

Continue Reading

Ans (h+g)/2g

*Plz do comment if ans is wrong*



225 Views · View Upvoters · Answer requested by Piyush Keshari

Upvote · 1

Share

 

Comment...

Promoted by IndiaRush

 

The way to stay trendy anywhere anytime.

Managing daily fashion in daily life can be a headache. Here's the only site that will make this easy.

Shop Now



Kuldip Singh Kalsi, Chief Chemist at PSPCL (1985-present)

Answered May 26

Continue Reading

h = ut +0.5 gt2

as u =0 and g =10 m/s2

Let total height = H

Total time = T

H = 0 5 gT2 = 5 T2

Distance travelled in nth second ( last second )

Sn = 0.5 g ( n2 — ( n-1)2 ) = 0.5 g ( 2n-1) = H/2

So 0.5 gT2 /2 = 0.5 g ( 2n -1) = 5 (2n -1 )

H/2 = 5 (2n-1)

Answer 2

HEY BUDDY HERE IS YOUR ANSWER BY

UDIT

the ball is released from the top of tower so
initial velocity,u =0

using second equation of motion

s= ut +1/2at^2

h/2=1/2gt^2 ( since u =0 and a =g)

h=gt^2

h=10×10^2

h=1000 m

thanks