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 thea. speed $$u$$b. distance of point of projection of ball I from base of tower (x)c. height of tower

Solution

## a. Time of ascent should be $$2s$$.so $$\dfrac{u\sin 60^o}{g}=2 \Rightarrow u=\dfrac{40}{\sqrt{3}}ms^{-1}$$b. $$x=(u\cos 60^o)t=\dfrac{40}{\sqrt{3}}\times\dfrac{1}{2}\times 2=\dfrac{40}{\sqrt{3}}m$$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 $$\dfrac{h}{x}=\tan 60^o$$$$\Rightarrow h=x\tan 60^o=\dfrac{40}{\sqrt{3}}\sqrt{3}=40m$$OR: height ascended by ball I is $$2s$$$$\Rightarrow h_1=u\tan 60^o\times 2-\dfrac{1}{2}10(2)^2=20m$$height descended by ball II in $$2s$$;$$h_2=\dfrac{1}{2}gt^2=\dfrac{1}{2}10(2)^2=20m$$Now $$h=h_1+h_2=20+20=40m$$Physics

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