NTSE Stage 2 SAT 2017 Question Paper

Question : 30 of 100

Marks: +1, -0

A ball released from rest at time t = 0 hits the ground. It rebounds in elastically with a velocity 5 m /

\s^{–1}

and reaches the top at t = 1.5 s, What is the net displacement of the ball from its initial position after 1.5s? (g = 10 m/

\s^{-2}

)

Solution:

\t_1

= time of fall

\t_2

= time of going up

\S_1

= initial height of fall

\S_2

= height acquired on going up

\S_2={\v^2-\u^2}/{2\a}

\;\;\;={0-25}/{2×(-10)}=1.25 \m

\t_2={0-5}/{-10}=0.5 \s

\t_1=1.5-0.5=1\s

\s_1=\ut_1+1/2\at_1^2

\;\;\;=0+1/2×10×1

\;\;\;=5\m

displacement

=5-1.25

\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\=3.75\m

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