Question

please tell the answers​

Answer 1

Explanation:

GIVEN:-

=>The object which is dropped in,

Height (h) = 120m

Acceleration (a) = 10ms^2

Initial velocity (u) = 0

TO FIND:-

=>The velocity and the distance travelled.

UNDERSTANDING THE CONCEPT:-

According to the question,

=> We know the height, acceleration and initial velocity. From this we can find the velocity for each seconds that id 1sec , 2secs, 3secs using first law of kinematics.

=> Then after finding the velocity, using third law of kinematics find the distance travelled.

$\huge\tt\fbox\purple{Answer}$

Using first law of kinematics,

For 1 second,

$\sf{}v = u + at$

$\sf{}v = 0 + 10 \times 1$

$\sf{}v = 10$

For 2 seconds,

$\sf{}v = 0 + 10 \times 2$

$\sf{}v = 20$

For 3 seconds,

$\sf{}v = 0 + 10 \times 3$

$\sf{}v = 30$

As it is given in ms^2,

The velocity will be 30^2 that is 900.

=>Now, using third law of kinematics,

$\sf{}v {}^{2} = u {}^{2} + 2as$

$\sf{}900 = 0 + 2 \times 10 \times s$

$\sf{}20s = 900$

$\sf{}s = 45m$

Therefore, The final velocity is 900 and the total distance travelled is 45m.