A stone dropped from the top of a building 1. How long does it take to fall 19.6m 2. How fast does it move at the end of this fall 3. What is its acceleration after 1sec and after 2 sec Given g=10m/s2
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1.
T = sqrt(2H/g)
= sqrt(2 * 19.6 / 10)
= 1.9 seconds
It takes 1.9 seconds to fall 19.6 m
2.
If 19.6 m is the height of the building then,
v = sqrt(2Hg)
= sqrt(2 * 19.6 * 10)
= 19.79 m/s
It moves with speed 19.79 m/s at the end of its fall.
3.
Acceleration is 10 m/s^2 through out its journey. Hence acceleration of stone after 1s and 2s is 10 m/s^2
[Note: If g is taken as 9.8 m/s^2 in (1) and (2) then answers of (1) and (2) will be 2s and 19.6 m/s respectively].
T = sqrt(2H/g)
= sqrt(2 * 19.6 / 10)
= 1.9 seconds
It takes 1.9 seconds to fall 19.6 m
2.
If 19.6 m is the height of the building then,
v = sqrt(2Hg)
= sqrt(2 * 19.6 * 10)
= 19.79 m/s
It moves with speed 19.79 m/s at the end of its fall.
3.
Acceleration is 10 m/s^2 through out its journey. Hence acceleration of stone after 1s and 2s is 10 m/s^2
[Note: If g is taken as 9.8 m/s^2 in (1) and (2) then answers of (1) and (2) will be 2s and 19.6 m/s respectively].
yumi1:
What does sqrt(2Hg) means.
nd how hv u calculated the time?
By tgis formula? V=u +gt
this*
It means v = √(2Hg). We got that by using v^2 - u^2 = 2aS. u = 0 for freely falling body.
Ohhh ...
thx a lot
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