From a tower of height ha particle is projected horizontally with velocity u and another thrown

Answer

Verified

Hint: the motion is in one dimension. No external force is acting on the system and the only force which acts is the gravitational force and the body moves both upside and down under the influence of gravitational force. Acceleration due to gravity is taken to be g.

Complete step by step answer:

The particle is thrown vertically upwards from the top of the tower and the height of the tower is H. Initial velocity=uFinal velocity=0Acceleration=-gUsing v=u+at0=u-gt\[{{t}_{1}}=\dfrac{u}{g}\]--------------(1)If \[{{t}_{2}}\] be the time taken to hit the ground then:\[-H=u{{t}_{2}}-\dfrac{1}{2}\times gt_{2}^{2}\]But it is given in the question that \[{{t}_{2}}=n{{t}_{1}}\]Putting this in above equation we get \[\begin{align} & -H=u\dfrac{nu}{g}-\dfrac{1}{2}\times g\dfrac{{{n}^{2}}{{u}^{2}}}{{{g}^{2}}} \\ & 2gH=n{{u}^{2}}(n-2) \\ \end{align}\]

So, the correct answer is “Option A”.

Additional Information:

Also when a body is thrown upwards then always at the topmost point the body takes a momentarily pause and when it starts to come back down towards the earth, in this case, the initial velocity is taken as zero and body strikes the earth back with some velocity.

Note:

This problem we have to keep in mind that any motion in an upward direction, the acceleration due to gravity is negative and in downward motion acceleration due to gravity is taken positively. All other values to be taken in standard SI units.

Text Solution

`v``2v``sqrt(2)v``v//2`

Answer : c

Solution : For the first body, `h=1/2"gt"^(2)........(i)` and `c=vt..........(ii)` From eqn. (i) and(ii) `h=1/2g.((x^(2))/(v^(2)))...........(iii)` Dividing eqn (iii) by eqn (iv) we get `1/2=(x^(2))/(v^(2))xx(v'^(2))/(4x^(2)) or v'=sqrt2v`

Text Solution

Solution : For the vertical motion AO, `h=0*t+1/2"gt"^2 or, t=sqrt((2h)/g)` As it corresponds to the actual motion AB , the time of flight is ,`t=sqrt((2h)/g)`.

A body of mass m is projected horizontally with a velocity v from the top of a tower of height h and it reaches the ground at a distance x from the foot of the tower. If a second body of mass 2 m is projected horizontally from the top of a tower of height 2 h, it reaches the ground at a distance 2 x from the foot of the tower. The horizontal velocity of the second body isVA. 1/2B. 2 VC. VD. √2 v

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Suggest Corrections