Question

An object is being thrown at a speed 20m/s in a direction 45 degree above the horizontal .The time taken by the object to return to the same level is 1) 20/g 2)20g 3)20root2/g 4)20root2g

Answer 1

The question asked is for projectile motion.

So, actually in a projectile motion the time of total flight is given by,

T = 2u*sin theta / g
where,
u is the initial velocity
Theta is the angle of projection.

This formula can be derived by using the 2nd equation of kinematics where you can make the displacement of the projectile zero and the take the components of the velocity and the acceleration along the y-axis.

Now, coming to the question now it remains a very simple one to solve.

Time of flight = 2*20*sin45°/g
= 40*1/√2 / g = 20√2 / g

Hence , the option 3) is absolutely correct for the above question.
Hope this helps you !!

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Answer 2

Answer:20root2/g

Explanation:T=2usina/g

u =20m/s

a=45degree

T=2×20×sin45/g

T=40×1/root2/g

=40 root2/2×g

=20 root2/g