A Stone is Thrown in a Vertically Upward Direction Problem

Q. A stone is thrown in a vertically upward direction with a velocity of 5 ms^-1. If the acceleration of the stone during its motion is 10 ms^-2 in the downward direction, what will be the height attained by the stone and how much time will it take to reach there?

Answer:- 

• u (Initial velocity) = 5 m/s

• v (Terminal velocity) = 0 m/s (At the maximum height, the stone will be at rest) 

• Acceleration would be = 10 ms^-2

• This acceleration would be in the direction opposite to that of the stone, which would be = – 10 ms^-2

Formula : According to the third equation motion, 

    v² = u² + 2as

Distance travelled by stone :

s = v² – u²/2a

 s = (0² – 5²)/ 2х10

S = 1.25 metres

According to the equation of motion, 

v = u + at 

Time taken to reach position of rest:

t = (v – u)/a

t = (0 – 5)/(-10) 

The Time taken = 0.5 seconds.

Now, we can say that in 0.5 seconds, the stone will reach max height, which would be 1.25 metres.