Question

A stone is dropped from the top of a building 147 m high how long will it take to reach the ground? What will be its velocity when it strikes the ground

Answer 1

Answer:

time taken by stone to reach ground

➪ 30 seconds

________________

velocity by which it hit the ground

➪ 294 m/s

Step by step explanations :

given that,

A stone is dropped from the top of a building 147 m

Here,

initial velocity of the stone = 0 m/s

[stone was at rest]

gravitational acceleration = -9.8 m/s²

let the time taken by the stone to reach the ground be t

here,

we have,

initial velocity(u) = 0 m/s

gravitational acceleration(g) = -9.8 m/s²

height(h) = -147 m

[height is downwards]

time taken(t) = t

by the gravitational equation of motion,

h = ut + ½ gt²

putting the values,

-147 = 0(t) + ½ × (-9.8)t²

-4.9t = -147

t = -147/-4.9

t = 30

so,

time taken by stone to reach ground

➪ 30 seconds

now,

let the velocity by which it hit the ground be v

v = u + gt

v = 0 + (-9.8)(30)

v = -294 m/s

negation of velocity shows the velocity downwards

so,

velocity by which it hit the ground

➪ 294 m/s

____________________

time taken by stone to reach ground

➪ 30 seconds

________________

velocity by which it hit the ground

➪ 294 m/s

Answer 2

$\underline{\mathfrak{Answer:-}}$

t = 5.4 sec

v = 53 m/s

$\underline{\mathfrak{Explanation:-}}$

Given

intial velocity (u) = 0

height (h) = 147m

Gravitational acceletation(g) =9.8 m/s²

To Find

Velocity of the stone and time taken by the stone to reach the ground

Solution

From the equation of motion

$\boxed{h = ut + \dfrac{1}{2}g{t}^{2}}$

On putting the values

$\mathsf{147 = (0)(t)+\dfrac{1}{2}(9.8){t}^{2}}$

$\mathsf{147 = 4.9{t}^{2}}$

$\mathsf{\dfrac{147}{4.9}= {t}^{2}}$

$\mathsf{{t}^{2}= 30}$

$\mathsf{{t}^{2}= 30}$

$\mathsf{t= \sqrt{30}}$

$\mathsf{t= 5.4 sec}$

Now,

$\boxed{v = u+gt}$

On putting values

$\mathsf{v = 0+(9.8)(5.4)}$

$\mathsf{v = (9.8)(5.4)}$

$\mathsf{v = 53 m/s (approx)}$