Question

Ball I is thrown towards a tower at an angle of ${60}^o$ with the horizontal with unknown speed (u). At the same moment, ball II is released from the top of tower as shown in Fig. Balls collide after 2 s and at the moment of collision, the velocity of ball I is horizontal. Find the
a. speed $u$
b. distance of point of projection of ball I from base of tower (x)
c. height of tower

Answer 1

a. Time of ascent should be $2s$.

so $\frac{u\sin {60}^o}{g}=2\Rightarrow u=\frac{40}{\sqrt{3}}m{s}^{-1}$

b. $x=\left(u\cos {60}^o\right)t=\frac{40}{\sqrt{3}}\times \frac{1}{2}\times 2=\frac{40}{\sqrt{3}}m$

c.The height of tower can be forund using of relative velocity. If the balls have collide, the initial velocity of ball I should be towards ball II.

For this $\frac{h}{x}=\tan {60}^o$

$\Rightarrow h=x\tan {60}^o=\frac{40}{\sqrt{3}}\sqrt{3}=40m$

OR: height ascended by ball I is $2s$

$\Rightarrow h_1=u\tan {60}^o\times 2-\frac{1}{2}10(2{)}^2=20m$

height descended by ball II in $2s$;

$h_2=\frac{1}{2}g{t}^2=\frac{1}{2}10(2{)}^2=20m$

Now $h=h_1+h_2=20+20=40m$