Question

Ab bol which is thrown vertically upward reaches the roof of a house is hundred metre high at the moment this ball is thrown vertically upward another ball is dropped from the rest vertically downward from the roof of the house at which the height with ball ball pass each other g=9.8

Answer 1

Judwaa 2 Star gold par aa rahi hai

dekh lo bhaiyon or behno

Answer 2

Answer:

Whenever an object is thrown vertically upwards as long as the object is on the Earth, a gravity force acts on it and yields weight. The definition of weight is

w=mgw=mg

Whatever the force which acts on an object, the law of motion is necessary to find position of the object at any time.

F=maF=ma

Define upwards is positive and downwards is negative. Thus the force ww is negative, since gravity points downwards.

−w=ma−w=ma

−mg=ma−mg=ma

a=−ga=−g

Our goal is to find position at any time x(t)x(t) . Since acceleration is the second derivative of position with respect to time, we get

d2xdt2=−gd2xdt2=−g

dxdt=v0−gtdxdt=v0−gt

x(t)=x0+v0t−12gt2x(t)=x0+v0t−12gt2

Assume that the object is thrown from origin. This reduces our initial condition

x0=0.x0=0.

Given from the question v0=3.5m/sv0=3.5m/s and let’s approximate g=10m/s2g=10m/s2 , now we have our position function and velocity function

x(t)=3.5t−5t2x(t)=3.5t−5t2

v(t)=3.5−10tv(t)=3.5−10t

As the ball is going upwards and reaches the peak, it temporarily stops at certain position and obviously its velocity is zero. Let’s call this time as tpeaktpeak . We utilize this fact by substituting v(tpeak)=0v(tpeak)=0 , or

v(tpeak)=3.5−10tv(tpeak)=3.5−10t

0=3.5−10tpeak0=3.5−10tpeak

tpeak=0.35stpeak=0.35s

Since this time is time when the ball reaches its peak, the height of the flight results from substituting tpeaktpeak to x(t)x(t) , or

x(tpeak)=3.5(0.35)−5(0.35)2x(tpeak)=3.5(0.35)−5(0.35)2

x(tpeak)=0.6125x(tpeak)=0.6125

So the highest position reached by the ball is 0.6125m0.6125m relative to position where the object is initially thrown with duration 0.35s.0.35s.