Question

A stone is thrown in a vertically upward direction with a velocity of 6 ms 1. If the acceleration of
the stone during its motion is 10 m s-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 1

Answer:

Maximum height = 1.8 m

time taken = 0.6 seconds

Step by step explanations :

given that,

A stone is thrown in a vertically upward direction with a velocity of 6 ms 1.

here,

initial velocity of the stone = 6 m/s

let the Maximum height attained by the stone be h

final velocity = 0

[stone will at rest]

and

time taken to reach the maximum height be t

given

the gravitational acceleration = 10m/s²

since acceleration is in downward direction

so,

acceleration = - 10 m/s²

now,

we have,

initial velocity(u) = 6 m/s

final velocity(v) = 0 m/s

gravitational acceleration(g) = - 10 m/s²

so,

by the gravitational equation of motion,

v² = u² + 2gh

putting the values,

(0)² = 6² + 2(-10)h

-20 h = -36

h = -36/-20

h = 1.8 m

also,

v = u + gt

again putting the values,

0 = 6 + (-10)t

-10t = -6

t = -6/-10

t = 0.6 s

so,

height attained by the stone

= 1.8 m

time taken to reach maximum height

0.6 seconds

Answer 2

Answer:

Explanation:

Given :-

Initial velocity of the stone (u) = 6 m/s  

Final velocity of the stone (v) = 0 m/s  

Acceleration by the stone (a) = -10 m/s²

To Find :-

1. Height attained by the stone, i.e. Distance, s =?  

2. Time (t) taken to reach the height = ?

Formula to be used :-

Motion's 3rd equation v² = u² + 2as

Motion's 1st equation v = u + at

Solution :-

1. Height attained by the stone, i.e. Distance

Putting all the values, we get

⇒ v² = u² + 2as

⇒  (0)² = 6² + 2(-10)s

⇒  -20 s = -36

⇒ s = -36/-20

⇒ s = 1.8 meters

2. Time (t) taken to reach the height

Putting all the values, we get

⇒ v = u + at

⇒ 0 = 6 + (-10)t

⇒  -10 t = -6

⇒  t = -6/-10

⇒  t = 0.6 seconds

Hence, the Time taken to reach the height is 0.6 seconds.