Question

If a particle takes t second less and acquires a velocity of V m/s more in falling through the same distance on two planets where the accelerations due to gravity are 2g and 8g respectively, then

(a) V =4gt

(b) V= 5gt

(c) V= 2gt

(d) V= 16gt​

Answer 1

Answer:

V = 4gt

Explanation:

If a particle takes t second less and acquires a velocity of V m/s more in falling through the same distance on two planets where the accelerations due to gravity are 2g and 8g respectively

a = 2g

Time taken = T

Distance = (1/2)2gT² = gT²

Velocity achieved = aT  = 2gT

a = 8g

Time taken = T - t

Distance = (1/2)8g(T-t)² = 4g(T-t)²

Velocity achieved = 8g(T - t)

8g(T - t) =  2gT + V

=> 6gT - 8gt = V   - eq 1

gT² = 4g(T-t)²

=> T² = 4(T - t)²

=> T = 2(T - t)

=> T = 2T - 2t

=> T = 2t

putting in eq 1

6g(2t) - 8gt = V

=> V = 4gt

Answer 2

Answer:

Explanation:

In first planet

a=2g

Time taken =T

Distance =21​2gT2=gT2

Velocity achieved =aT=2gT

In second planet

a=8g

Time taken =T−t

Distance =21​8g(T−t)2=4g(T−t)2

Velocity achieved =8g(T−t)

8g(T−t)=2gT+V

⟹6gT−8gt=V            ........(1)

Since the distance is same,

gT2=4g(T−t)2

⟹T2=4(T−t)2

⟹T=2(T−t)

⟹T=2T−2t

⟹T=2t

Putting the value of T in equation (1), we get

6g(2t)−8gt=V

⟹V=4gt