A man drops a ball downside from the roof of a tower of height 400 m. At the same time another ball is thrown upside with speed of $50 {ms}^{- 1}$ , they will meet at which height from the base of the tower:

see full answer

High-Paying Jobs That Even AI Can’t Replace — Through JEE/NEET

🎯 Hear from the experts why preparing for JEE/NEET today sets you up for future-proof, high-income careers tomorrow.

An Intiative by Sri Chaitanya

80 m

240 m

100 m

320 m

answer is C.

(Unlock A.I Detailed Solution for FREE)

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

Let both balls meet at point P after time t

The distance travelled by ball A, $h_{1} = \frac{1}{2} g t^{2}$

The distance travelled by ball B, $h_{2} = u t - \frac{1}{2} g t^{2}$

$h_{1} + h_{2} = 400 m \Rightarrow u t = 400, t$

= 400/50 = 8 sec

$h_{1} = 320 m and h_{2} = 80 m$

Watch 3-min video & get full concept clarity

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test