Question

a stone is thrown vertically with velocity 20 m/s. how much time it takes to reach ground (g=10m/s)​

Answer 1

(Consider g=10ms2 for ease of calculation). So it reaches 20 m from the peak of tower of height 25 m. So to reach to ground,the stone has to cover: s=20+25=45 m.

Answer 2

Given :-

• Initial Velocity of stone = u = 20 ms-¹

• Acceleration due to gravity = g = 10 ms-²

Concept :-

When a body is thrown upward it has initial velocity only.

v = u - gt

As gravity is acting opposite.

0 = 20 - 10t

10t = 20

t = 2 s

Total distance covered,

S = ut - 1/2 gt²

S = 20 × 2 - 1/2 × 10 × 2 × 2

S = 40 - 20

S = 20 m

This is the maximum distance covered while going for a round trip; The total distance = D = 2S = 40 m

Using second equation of motion,

40 = 20t - 1/2 × 10t²

5t² - 20t + 40 = 0

t² - 4t + 8 = 0

D = b² - 4ac

D = 16 - 32

D = -16

t = -b±√D/2a

t = 4 + 4/2×1

t = 4s

OR,

t = 4 - 4/2×1

t = 0 s (Not valid)

Hence,

Time taken = t = 4s