Question

th. From the
e same velocity
A stone dropped from a building of height hand it reaches after t seconds on earth. From
building if two stones are thrown (one upwards and other downwards) with the same
they reach the earth surface aftert, and t, seconds respectively, then
(1) t= t, -tz
(2) t = 472
(3) t= ſtata
(4) t= 1212
th
A particle is projected vertically upwards from ground level with a speed of 50 m/s. For how long w
more than 70 m above the ground ? (g = 10 m/s2)
If a body travels half of total path in the last second of its fall from rest, then find:
(a) the time of fall and (b) the height of its fall. (g = 9.8 m/s2)
A ball is projected vertically up with an initial speed of 20 m/s and acceleration due to gravity is 10 mis
(a) How long does it take to reach the highest point ?
(b) How high does it rise above the point of projection?
(c) How long will it take for the ball to reach a point 10 m above the point of projection second time?
A body is released from a height and falls freely towards the earth, exactly 1second later another too
released. What is the distance between the two bodies 2 second after the release of the second boaj
= 9.8 m/s).​

Answer 1

Answer:

answer of bottom one

Body #1

d = 1/2 at^2

so after 1 sec, d = 4.9m After 2 sec, d = 19.6m. After 3 sec, d = 44.1m

Body #2

Same as for body 1….after 2 seconds, it has fallen 19.6m.

Meanwhile, Body #1 has fallen, as I showed above, 44.1m.

So the distance is 44.1 - 19.6 = 24.5m

Explanation:

body travel half distance answer

The first approach to this question is to define the variables. So , let's do that:

Let the total distance from the ground be “h” , so the distance travelled in last one second will be “h/2”.

So as it is free fall , u(initial velocity) = 0

Let's take the velocity of the object after traveling half the distance as “v” and the velocity with which it hits the ground as “v'”

And as always acceleration = 9.8 m/s/s (As you'll see “g” doesn't matter , it cancels out.)

Also time taken to complete first half is taken as “t” and to complete second half is “t'” = 1s.

NOTE : I am gonna take the Equations of motion as they are , if you don't know them please ask Google for derivations.

Now ,

First let's calculate v(velocity after half of the distance has been travelled)

v=u+at(Equationofmotion)v=u+at(Equationofmotion)

(As u = 0 ) the above equation becomes

v=at−−−−−(I)v=at−−−−−(I)

Now , with that out of the way.

Another equation of motion is as you know

S=ut+1/2(at2)(EquationofMotion)S=ut+1/2(at2)(EquationofMotion)

So , as the first and the last half distance travelled is h/2 , we can achieve two more equations specific to our question from the above equation.

h/2=0∗t+1/2(at2h/2=0∗t+1/2(at2)−−−−−−−(II))−−−−−−−(II)(As S = h/2 and u = 0)

Also , for the last second travelled

h/2=ut+1/2(a∗1)−−−−−−−−−(III)h/2=ut+1/2(a∗1)−−−−−−−−−(III) (As t = 1s and 1^2 = 1)

Now remember equation (I) above , yeah I do too ,

Let's replace u of the equation (III) with the v from equation (I) cause initial velocity of the last second travelled is v as defined in the variables.

Now substituting ,

h/2=(at)(t′)+1/2(a)h/2=(at)(t′)+1/2(a) (As v = at)

The Equation becomes ,

h/2=(at)+1/2(a)−−−−−(IV)h/2=(at)+1/2(a)−−−−−(IV) (As t' = 1)

So , now let's equate equation (II) and equation (IV)

(at)+a/2=1/2(at2)(at)+a/2=1/2(at2)

=>at+a/2=(at2)/2=>at+a/2=(at2)/2

=>a(t+1/2)=a(t2)/2=>a(t+1/2)=a(t2)/2

=>t+1/2=t2/2=>t+1/2=t2/2 {As a got cancelled , I told you :)}

=>2t+1=t2=>2t+1=t2 (Taking 2 in the denominator to the other side)

=>t2−2t+1=0=>t2−2t+1=0

Now we have got a quadratic equation for “t” , I assume you know how to solve for “t”.

Sorry but my teacher says never assume sh*t , so I am gonna do it anyway.

rest are in attachment