A stone is thrown with some initial speed vertically downwards from a tower of height 40 m. It strikes the ground in 2 s. What is the initial speed of the stone? (Take g = 10 m/s²)

(1) 5 m/s
(2) 15 m/s
(3) 20 m/s
(4) 10 m/s

To solve this problem, we will use the equations of motion under constant acceleration due to gravity. The equation we will use is:
s = u t + 2 1 ​ a t 2
where:

s is the displacement (in meters), which is the height of the tower in this case, 40 m .
u is the initial speed (in meters per second) that we are solving for.
t is the time (in seconds) it takes for the stone to hit the ground, which is 2 s .
a is the acceleration due to gravity, 10 m/s 2 .

Substituting these values into the equation, we have:
40 = u × 2 + 2 1 ​ × 10 × ( 2 ) 2
Simplify and solve for u :

First, calculate 2 1 ​ × 10 × 4 = 20 .
So the equation becomes: 40 = 2 u + 20
Subtract 20 from both sides: 20 = 2 u
Divide both sides by 2: u = 10 m/s

Therefore, the initial speed of the stone is 10 m/s .
The correct multiple-choice option is (4) 10 m/s .

Answered by OliviaMariThompson | 2025-07-22

The initial speed of the stone thrown downward from a height of 40 m, striking the ground in 2 seconds, is calculated to be 10 m/s. Therefore, the correct option is (4) 10 m/s.
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Answered by OliviaMariThompson | 2025-07-23