Question

an object is dropped from a height of 80m. it will remain in air for time?(take g=10m/s2

Answer 1

using the third eq of motion v^2-u^2=2gh
 v^2=2*10*80
v^2=1600
v=40m/s
t=v-u/g
40-0/10
4seconds

Answer 2

Answer:

Time, t = 4 seconds

Explanation:

It is given that,

The height from which the object is thrown, h = 80 m

Initial speed of the object, u = 0

It is moved under the action of gravity. We can use the second equation of motion as :

$h=ut+\dfrac{1}{2}gt^2$

$t=\sqrt{\dfrac{2h}{g}}$

$t=\sqrt{\dfrac{2\times 80}{10}}$

t = 4 seconds

So, the ball will in air for 4 seconds. Hence, this is the required solution.