## Tuesday, March 24, 2009

### Solution to Q.1. - Go for it

What's the question

If we draw the trajectories of the two shells, firing from the same point at 60 deg and 45 deg, we find that there can only be one collision point as shown. Now we have to find out after how much time the second shell (at 45 deg) has to be fired so that the shells collide, ie reach the point of collision at the same time.
u= 250 m/s (given)
Plotting the trajectories in an X-Y plane:
Let the shells be fired from origin O (0,0).
Let the shells collide at (x,y).
Let shell 1 at 60deg be fired at time = 0.
Let shell 2 at 45 deg be fired after t sec , ie at time =t.
Let the collision take place at time = T
• Shell 1 took T sec to reach (x,y) => shell1’s horizontal component took T sec to travel x displacement and vertical component took T sec for y displacement    (Components?How?Click here)
i.e x = u cos60 . T ,             y = u sin60. T – (½)gT^2                how?
• Shell 2 took (T-t) sec to reach (x,y) => shell2’s horizontal component took T-t sec to travel x displacement and vertical component took T-t sec for y displacement
i.e x = u cos45 x (T-t) ,         y = u sin45 .(T-t) – (½)g (T-t)^2    how?

Eliminating x and y we get
u cos60 . T = = u cos45 x (T-t) ---eqn(1) and
u sin60. T – (½)gT^2 = u sin45 .(T-t) – (½)g (T-t)^2 ----eqn(2)

eqn(1) => T = t.cos45 / ( cos45 – cos60) = sqrt(2) . t / ( sqrt(2) - 1 )

substituting the value of T in eqn(2) we get
t = 2u sin(60-45) / [ g(cos60 + cos45)] = 11 sec. (answer)

* sqrt(2) => square root of 2

#### 1 comment:

1. can we do this with the help of relative motion...