Question

From a tower of height H, a particle is thrown vertically upwards with a speed u. The time taken by the
particle, to hit the ground, is a n times that taken by it to reach the highest point of its path.
The relation between H, u and n is:
(JEE(Main)-2014; 4

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Answer 1

Answer:

-20H = nu²

Explanation:

When we consider the motion where the particle reaches max. height from the point of projection:-

Initial Velocity = u

Final velocity = 0

by 1st equation of mtotion,

v = u + at

⇒ 0 = u -gt

⇒ t = u/g

When we consider the motion where the particle reaches the ground from the point of projection,

Initial Velocity = u

Final velocity = v

Time taken = T = n*t = n (u/g)

By 2nd equation of motion,

s = ut + 1/2 at²

⇒ -H = u T - 1/2 g T²

⇒ -H = nu²/g - nu²/2g

⇒ -H = nu²/2g

⇒ -2gH = nu²

Assuming g = 10m/s²

⇒ -20H = nu²

Answer 2

Answer:

Let us begin with the first equation, v=u+at. This equation only talks about the acceleration, time, the initial and the final velocity. Let us assume a body that has a mass “m” and initial velocity “u”. Let after time “t” its final velocity becomes “v” due to uniform acceleration “a”. Now we know that:

Acceleration = Change in velocity/Time Taken

Therefore, Acceleration = (Final Velocity-Initial Velocity) / Time Taken

Hence, a = v-u /t or at = v-u

Therefore, we have: v = u + at

v² = u² + 2as

We have, v = u + at. Hence, we can write t = (v-u)/a

Also, we know that, Distance = average velocity × Time

Therefore, for constant acceleration we can write: Average velocity = (final velocity + initial velocty)/2 = (v+u)/2

Hence, Distance (s) = [(v+u)/2] × [(v-u)/a]

or s = (v² – u²)/2a

or 2as = v² – u²

or v² = u² + 2as

s = ut + ½at²