General Knowledge - Physics - Discussion

Velocity:600*5/18 m./sec.=500/3m./sec. Hy=Uy*t+1/2g*t*t : 1960=0+ 1/2*9.8*t*t therefore t=20 sec. So Range AB=v*t=500/3*20 m.=10/3 km.=3.33 km.

What is projectile motion?

V=166.66m/s
Ht=1960m
t=20sec
R=v*t=166.66*20=3.33km.

Where did you get the value of "t"?

Actually what is this question talking about when the time is not given.

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.

Could anyone explain this please?

How come the time is 20 Seconds?

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.