Question

34. A particle is thrown vertically up with speed u so
that distance covered in last second of flight is
35 m. If g = 10 m/s2 then initial speed of throw is
(1) 20 m/s
(2) 30 m/s
(3) 40 m/s
(4) 50 m/s​

Answer 1

answer : option (3) 40 m/s

explanation : 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

$S_n=u'+\frac{1}{2}g(2n-1)$

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.