Question

For a particle projected vertically upwards under gravity travels equal distance during 5th and 6th second of its motion. Find its projection speed (g = 9.8 m/s2)



(1) 50 m/s



(2) 30 m/s



(3) 49 m/s



(4) 29.4 m/s

Answer 1

3.49 m/s ... ... ...... .....

Answer 2

A particle is projected vertically upwards with speed v m/s.

so, distance travelled by particle in nth second is given by,

$S_n=v-\frac{g}{2}(2n-1)$

a/c to question,

distance travelled in 5th second = distance travelled in 6th second

but we see , v - g/2 (2 × 5 -1) ≠ v - g/2(2 × 6 - 1)

it means, motion of particle is shown as figure.

distance travelled in 5th second = v - g/2(2 × 5 - 1)

= v - 9g/2

distance travelled in 6th second = 1/2 g(1)² = g/2

[ we can assume particle fall from a height h, then distance travelled by particle in 6th second = distance travelled by particle in 1st second falling from certain height ,h ]

now, v - 9g/2 = g/2

v = (9 + 1)g/2 = 5g = 50m/s

hence, projection speed = 50m/s.