Question

a body released from the top of a tower of height h takes t second to reach the ground at time 1/2t the height of the body from the ground is?

Answer 1

4.9 unless i using wrong formula
D=.5 * accelaration * time squared

.5*9.8

Answer 2

the ans=3h/4

Answer 3

4.9 meters
All objects fall at same 9.8 m/s^2

Answer 4

look like typical question

Answer 5

3h/4 (75% of the initial height) is right

Answer 6

first find t in terms of h. For the stone to drop from rest until it reaches the ground : we use
\[s= ut + 0.5at^2\]
where s=h, u =0, a assumed to be 10
then you will get
\[t=\sqrt{h/5}\]
After 1/2 t , let the stone travels a distance of x unit
still use the formula \[s= ut + 0.5at^2\]
\[x= 0 + 0.5(10)(0.5t)^2\]
\[x= 5/4 t^2\]
sub the value of t you found just now and you will get x=h/4
so it's height from the ground is = h-h/4 = 3h/4

Answer 7

should be \[x=(5/4)t^2\]

Answer 8

I figured :)

Answer 9

how 5/4???

Answer 10

not getting that...

Answer 11

\[x= (0.5)(10)(0.5^2)t^2 \]
use your calculator to get the value 5/4

Answer 12

i didnt see 0.5^2

Answer 13

thanzzz