Question

a body is thrown with velocity v0 from the height H its velocity after 1 sec is ?

Answer 1

Hint: First we need to draw the diagram according to the question. Now we know the formula for displacement of a body with respect to acceleration and time. We need to replace the values of acceleration and height and time to get the required equation.

Complete step by step answer:

Here we see a body is thrown from the ground with initial velocity ‘u’, the maximum height that the body reaches is ‘h’, we know at the maximum height the final velocity ‘v’ must be zero, and the body reaches the maximum height in ‘T’ seconds. After reaching the maximum height the body starts to fall down with an acceleration that is the same as the acceleration due to gravity.

Now we can say that,

h=ut−12gt2 ,when we are considering motion from ground to maximum height.

As we know the height it reached is ‘T’, then

h=uT−12gT2.

Now if we consider option C, the option says that, at a time 2T its velocity is – u

If the body goes up in time ‘T’ then it must come down in time ‘T’ with the same velocity and acceleration, because the velocity and acceleration is constant throughout the system.

Now if the velocity of the body is the same and now when the body is coming down from its maximum height its velocity is considered as a negative velocity as the direction is opposite.

So, the body has the same magnitude of velocity as when it was released from the ground, but it has just the opposite direction, so we can write that as – u.

Hence, the correct answer is option C.

Note:

In the formula h=ut−12gt2 when we are considering motion from ground to maximum height, the negative sign is because the body is opposing the motion of gravity. We know that the actual formula is S=ut−12at2 , but here we know the distance and we also know that the acceleration working is due to gravity, hence we can say that a = g and S = h.

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