Question

A particle is thrown vertically up with speed u so that distance travelled in last

Answer 1

Answer:

particle is thrown vertically up with speed u m/s. distance covered in the last second of flight is 35m.

time taken to reach the maximum height , t = (u/g)sec,

particle covers 35m in last time of flight. means, it covers 35m before striking the ground.

in descending , initial velocity = velocity at maximum height = 0

now applying formula,

distance travelled in nth second

here, Sn = 35, g = 10m/s², n = t = (u/g) , initial velocity , u' = 0

so, -35 = 0 + 1/2 × (-10) × [2(u/g) - 1]

or, 35 = 10(u/10) - 5

or, 40 = u

or, u = 40 m/s

hence, initial velocity of particle is 40m/s.

Answer 2

Answer:

distance travelled in last second

= u + 1/2 (g) (2n-1)th

here nth is second

and g is negative