A ball is dropped from the top of a tower. After 2s another ball thrown vertically downwards with a speed of 40ms−1, After how much time and at what distance below the the top of tower the balls meet?
Open in App
Verified by Toppr
If 2nd ball takes t time before meeting . ball 1 has traveled for (t+2)s Both balls traveled same dist. ∴∣dfrac12(t+2)2=ut+21g(t)2 21×10(t+2)2=40t+2t×10(t)2 5(t+2)2=40t+562 5(t2+4+4t)=40t+5t2 h=21gt2 for 1st ball h=40(t−2)+21g(t−2)22nd ball 21gt2=40(t−2)+21(t−2)2 40(t−2)=21g(t2−(t−2))2 4t−8=2t−2⇒t=3 h=21gt2 =21×10×(3)2=45m