Calculate the distance traveled by a body dropped from rest in $0.1 s$ . Take $g = 10 N k g^{- 1}$

50 m

No worries! We‘ve got your back. Try BYJU‘S free classes today!

5 m

No worries! We‘ve got your back. Try BYJU‘S free classes today!

0.05 m

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

0.5 m

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C $0.05 m$
Initial speed of the body $u = 0 m / s$
Distance travelled in $0.1$ s is given by $S = u t + \frac{1}{2} \times g t^{2}$
where $t = 0.1$ s
$∴$ $S = 0 + \frac{1}{2} \times 10 \times (0.1) (0.1)$
$⟹ S = 5 \times 10^{- 2} m = 0.05 m$

Suggest Corrections

Similar questions

Q. A block starts accelerating at

5 m / s^{2}

from rest on a frictionless surface. Calculate the distance traveled by the block in first 4 seconds?

A body starts from rest and falls vertically from a height of 19.6 meters. If g=9.8 m/sec2. Then, what is the distance traveled by the body in the last 0.1 second of its motion?

Q. A body of mass

100 g

is at rest on a smooth surface. A force of

0.1 N

starts to act on it for

5 s

. Calculate the distance travelled by the body during this time interval.

Two masses 90 kg and 160 kg are at a distance 5 m apart. Compute the magnitude of intensity of the gravitational field at a point distance 3m from the 90 kg and 4m from the 160 kg mass. G=6.67 $\times$ $10^{- 11}$ SI units..

A body dropped from a tower covers 25m in the last second of its fall. Then, the height of tower is [ take g= 10 m/s^2]

(1) 45m (2) 20m (3) 40m (4) 50m

Join BYJU'S Learning Program

Calculate the distance traveled by a body dropped from rest in 0.1 s. Take g=10 Nkg−1

The correct option is C 0.05 mInitial speed of the body u=0m/sDistance travelled in 0.1 s is given by S=ut+12×gt2where t=0.1 s∴ S=0+12×10×(0.1)(0.1)⟹S=5×10−2 m=0.05 m

Calculate the distance traveled by a body dropped from rest in $0.1 s$ . Take $g = 10 N k g^{- 1}$

The correct option is C $0.05 m$
Initial speed of the body $u = 0 m / s$
Distance travelled in $0.1$ s is given by $S = u t + \frac{1}{2} \times g t^{2}$
where $t = 0.1$ s
$∴$ $S = 0 + \frac{1}{2} \times 10 \times (0.1) (0.1)$
$⟹ S = 5 \times 10^{- 2} m = 0.05 m$