A stone is dropped from a height of 100 m from the ground. Take accelaration due to gravity as $9.8 m / s^{2}$ .
Calculate:
(a) The velocity with which it reaches the ground.
(b) The time taken to reach the ground. [3 MARKS]

Open in App

Solution

Finding Velocity : 1.5 Marks
Finding Time : 1.5 Marks

(a) Here,
$u = 0, s = 100 m, a = 9.8 m / s^{2}$
$v^{2} = u^{2} + 2 a s$
$v^{2} = 0 + 2 \times - 9.8 \times - 100$
$v^{2} = 1960$
$v = 44.3 m / s$

(b) $v = u + a t$
$44.3 = 0 + 9.8 t$
$t = \frac{44.3}{9.8} = 4.52 s$

Suggest Corrections

Similar questions

A body is thrown vertically upwards with an initial velocity of 19.6 m/s. If accelaration due to gravity $g = 9.8 m / s^{2}$ . Calculate the following:
(i) The maximum height attained by the body.
(ii) The time to reach the ground. [5 MARKS]

Q. A body dropped from a height reaches the ground in

5 s

. The velocity with which it reaches the ground is (

g = 9.8 m s^{- 2}

A stone is dropped from a rising balloon at a height of 76 m above the

ground and it reaches the ground in 6s. What was the velocity of the

balloon when the stone was dropped? Take g=10 m/ $s^{2}$ .

Q. A stone is dropped from the hill at the height of 180 meters. Calculate the time to reach the ground assume acceleration due to gravity is 10

m / s^{2}

and ignore air resistance.

A body is thrown vertically up with an initial velocity of 60 m/s. If acceleration due to gravity g = $10 m / s^{2}$ , what is the time taken to reach a height of 100 m. [3 MARKS]

Join BYJU'S Learning Program

A stone is dropped from a height of 100 m from the ground. Take accelaration due to gravity as 9.8 m/s2. Calculate: (a) The velocity with which it reaches the ground. (b) The time taken to reach the ground. [3 MARKS]

Finding Velocity : 1.5 Marks Finding Time : 1.5 Marks (a) Here, u=0, s=100 m, a=9.8 m/s2 v2=u2+2as v2=0+2×−9.8×−100 v2=1960 v=44.3 m/s (b) v=u+at 44.3=0+9.8t t=44.39.8=4.52 s

A stone is dropped from a height of 100 m from the ground. Take accelaration due to gravity as $9.8 m / s^{2}$ .
Calculate:
(a) The velocity with which it reaches the ground.
(b) The time taken to reach the ground. [3 MARKS]

Finding Velocity : 1.5 Marks
Finding Time : 1.5 Marks

(a) Here,
$u = 0, s = 100 m, a = 9.8 m / s^{2}$
$v^{2} = u^{2} + 2 a s$
$v^{2} = 0 + 2 \times - 9.8 \times - 100$
$v^{2} = 1960$
$v = 44.3 m / s$

(b) $v = u + a t$
$44.3 = 0 + 9.8 t$
$t = \frac{44.3}{9.8} = 4.52 s$