General Knowledge - Physics - Discussion

BOMB WILL FALL UNDER THE EFFECT OF EARTH GRAVITY(9.8M/SEC SQUIRE) AND AEROPLANE WILL MOVE IN HORIZONTAL DIRECTION WITH ITS HORIZONTAL SPEED V(V=u+a*t).

Here the value of V will be 1600 KmpH(=166.66m/s) because its initial speed(u) is zero. The taken by bomb to reach on earth will be calculated by formula (S=ut+0.5at*t); here the S is the height and a is equal to earth gravity(g=9.8) and u=0, so formula will be H=0.5gt*t and thus value of t( time taken by bomb to reach on earth will be) will be 20 sec.

Now the bomb will move in the horizontal direction in 20 sec and with the speed( same speed of aeroplane) of 1600(=166.66m/s) and it will move in horizontal direction (distance = speed x time) by 166.66x20 sec.

Convert velocity in horizontal direction, Ux into m/s:
Ux = 600x1000/3600 = 500/3 m/s.
Velocity of the bomb during the flight is constant due to inertia. Therefore,
Initial velocity in vertical direction, Uy = 0,
Height = 1960 m.

Step 2: Find time 't' using the equation [s = ut + 1/2at^2]
Replace distance 's' by height 'h' and acceleration 'a' by gravity 'g'
h = 1/2gt^2
t = √(2h)/g = √(2x 1960/9.8) = 20 sec.

Step 3: Find the distance traversed by the bomb in horizontal direction:
Distance AB = vx t = 500/3 x 20 = 10/3 x 10^3m = 3.333 km.

v = 166.66m/s, h = 1960m.

Now s = u*t + 0.5*a*t*t which we have studied in Physics where, s = distance; u = initial velocity; a = acceleration; t = time.

In this context a = g(acceleration due to gravity).

Substituting given values in the previous equation we have,

1960 = 0+0.5*9.8*t*t(Since g=9.8m/s^2 and initial velocity,u=0).

Solving for t we get t = 20 seconds.

Now Distance = Velocity*time = 166.66(m/s)*20(seconds) = 3333.4m = 3.334km.

Distance = initial velocity *time + 1/2(acceration due to gravity * time^2).
i.e s = ut +1/2(gt^2).

For a freely falling bomb u = 0 So,

s = 1/2gt^2.

t = sqrt(2s/g) = sqrt(2*1960/9.8) = 20 seconds = 20/2600 hr.

Not the time during which bomb touches ground from the time when it was released is t and during that that time aeroplane will move a distance AB = speed of plane * time elapsed.

AB = (600)* 20/3600 = 3.33 km.

Given data-
velocity (v) = 600km/h = (600x1000)/3600.
v = 166.66m/s.

Height (h) = 1960m.

We know that, the second law of motion.
s = ut + 0.5 * a *t * t----> 1

Put the given values in equation (1)
1960 = (0xt)+(0.5xtxt).

t(square) = 1960/0.5
t = 20 sec.

But again we know that velocity = distance/time.
v = d/t,
d = (v) * (t),
d = 20 * 166.66,
d = 3333.2m,
d = 3333.2/1000 = 3.33 km.
Ans. (d) = 3.33 km.

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First. We will change m/s.
Then 600 * 1000/3600 = 166.666m/s.
Next. H=1/2at^2.so 1960=1/2*9.8*t^2.
Then t=20s.
Finally, Use H=ut equation.then H=166.66 * 20.
So 3333.333metres.
So answer 3.33km.

Solution: see in the question its given:
Velocity= 600Km/hr = 600x1000/3600 m/s=166.66m/s
Ht=1960m.

To Find: The range of the projectile (R).
Its actually a Projetile motion .

For a projectile Ht=1/2*g*t*t.There fore we get t= 20s.
R=V*t=166.66*20=3333.33m/s=3.333Km/hr.

SOLUTION:

Velocity = 600km/hr=600*1000/3600m/s.
= 166.66m/s.

FORMULA:

Ht = 0.5*a*t*t.
Ht = 0.5*9.8*2*2.
Ht = 19.6+0.4.
H t= 20 seconds.

R = V*t.
R = 166.66*20 SEC.

R = 333.2km.
R = 333.2/1000.
R = 3.33km.

Therefore R is 3.33km.

When body travels under the effect of gravity and when initial velocity g are not at 180 to each other i.e opposite then projectile motion in which the body follows a curved path.