Question

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Answer 1

Question:- A ball is thrown vertically upward with a given velocity 'u' such that it rises for T seconds. What is the distance travelled by the ball during the last one second of Ascent (in metres) (acceleration due to gravity in metre per second square)?

Solution:-

When the ball is thrown upward

initial velocity of the ball=u

time taken=T

acceleration=g

Using third law of uniformly accelerated motion

we get

S(total distance travelled)=uT+1/2 ×gT^2

to find distance travelled by the ball in (T-1) time, replace T by (T-1)

S'=u(T-1). +1/2 ×g×(T-1)^2

distance travelled during the last one second=S-S'

=uT+1/2 ×g×T^2 -u(T-1)-1/2×g(T-1)^2

=uT+gT^2/2-uT+u-1/2 ×g(T-1)^2

=gT^2/2-gT^2/2+u-u+g/2

=g/2

Hence it will cover the distance of g/2 during last one second of ascent....{hope it helps you}