If a body covers 36% of its height in last sec while falling it free from the top of the tower. What is the height if that tower?
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heya......
Let H be the height of the building and t be the time to fall through this height H.
Then H = ½ g t^2 = 5 t^2. ----------------- (1)
In (t-1) second the distance traveled is 1 - 0.36 H = 0.64 H.
0.64 H = 5 (t-1)^2.---------------------------(2)
(2) / (1) gives 0.64 = (t-1)^2/ t^2
Or 0.8 = (t-1) / t
Solving we get t= 5 second.
Using (1) we get H = 125 m.
ysm.....@kundan
Let H be the height of the building and t be the time to fall through this height H.
Then H = ½ g t^2 = 5 t^2. ----------------- (1)
In (t-1) second the distance traveled is 1 - 0.36 H = 0.64 H.
0.64 H = 5 (t-1)^2.---------------------------(2)
(2) / (1) gives 0.64 = (t-1)^2/ t^2
Or 0.8 = (t-1) / t
Solving we get t= 5 second.
Using (1) we get H = 125 m.
ysm.....@kundan
Rishavlal:
Bro send a photo in which you are making this ques. Yourself in copy briefly
don't worry, I will provide you. He just copied it from here.
https://www.nasa.gov/audience/forstudents/k-4/stories/nasa-knows/what-is-a-black-hole-k4.html
https://www.askiitians.com/forums/General-Physics/a-body-falls-freely-from-the-top-of-a-tower-it-c_184579.htm
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here equations of motion will help
S=ut+0.5at²
in case of free fall u=0
so S=0.5at².............(i)
Now in laste second it covers 36% of the total height
that means in (t-1)s it covers 64%of the height
substituting in equation we get
0.64S=0.5a(t-1)².................(ii)
Now devide eq(ii) from eq (i)
0.64=(t-1)²/t²
t-1=0.8t
0.2t=1
t=5s
acc was g=10m/s²
so
S=0.5x10x5²=5³=125m
S=ut+0.5at²
in case of free fall u=0
so S=0.5at².............(i)
Now in laste second it covers 36% of the total height
that means in (t-1)s it covers 64%of the height
substituting in equation we get
0.64S=0.5a(t-1)².................(ii)
Now devide eq(ii) from eq (i)
0.64=(t-1)²/t²
t-1=0.8t
0.2t=1
t=5s
acc was g=10m/s²
so
S=0.5x10x5²=5³=125m
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