Question

A ball is thrown vertically upwards with a given velocity 'u' such that it rises
for T seconds (T > 1),
what is the distance traversed by the ball during the last one second of ascent (in meters) ?
(Acceleration due to gravity is g m/s.)
plzzzzzzzzzz answer​

Answer 1

SOLUTION

Velocity at the toppest point (t) is zero

therefore v= u-gt

=)0 = u-gt

=) u= gt

=) t= u/g

velocity at time t-1

=) v= u-g(t-1)

=) v= gt-gt+ g

=) v=g

Distance travelled in last one second (t=1)

=)s= (velocity at time t-1) t- 1/2gt^2

=) g×1- 1/2× g× 1^2

=) g/2

hope it helps

Answer 2

Explanation

travelled distance in T sec

H1=u*T - ½g T². ------(1.

and travelled distance in( T-1) sec

H2= u*(T-1) - ½g*(T-1) --- (2.

Now travelled distance by ball in last second

=traveling distance in[ (T sec)- (T-1) sec]

Now subtracting equation 2 from 1 we get

= u - g/2 *( 2T -1)

Distance traveled in last sec

u - g/2 *( 2T -1)...

Hope you would like it...