Question

a body is projected vertically upwards with a speed 40m/s the distance travelled travelled by body in its last sec of its journey is​

Answer 1

Answer:

v^2 = u^2 - 2as

=> u = 0 body is at rest

=> v= 40 m/s

=> a= -10 m/s (g rounded to 10)

(u is cancelled out, since it is 0)

=> 40^2 = -2 * -10*s

= 1600 = 20s

=> s = 1600/20 = 80

=> distance travelled = 80 m

(ask any doubt you found in my explanation)

Answer 2

Answer:

The distance travelled by a body in the last second of its upward motion=

distance travelled by the body in the first second of its downward motion

( Please note that this point makes such questions very easy.)

Therefore, considering the body as a freely falling body,

s= 1/2gt*2( where t= 1s )

s= 5 m

Alternatively, we can use s= u- g(n- 1/2) which is the formula for distance covered in the nth second.

n in this case is the last second of upward journey.

Therefore, time taken for upward motion= u/g= 40/10= 4s

s= 40- 10( 4-1/2)

Or s= 40- 35

Or s= 5m