Question

A ball is dropped from a height of 80 m. The distance travelled by it in the fourth second will be​

Answer 1

Answer:

Given,

Height of building = s = 80m

Here,

v = final velocity

u = initial velocity

t = time

a = acceleration

s = distance

For velocity of ball after dropping it :

Here, initial velocity = u = 0 m/sec

We know that,

(v)^2 = (u)^2 + 2as

=> v^2 = (0)^2 + 2 × 10 × 80

=> v^2 = 1600

=> v = √1600 = 40 m/sec

Hence, velocity of ball till first collision = 40 m/sec

Let the distance covered by ball after first collision be 'x'. Then,

(0) = 400 - 2 × 10 × x

=> 20x = 400

=> x = 400/20 = 20 m

Hence, distance covered by ball after first collision with floor = 20 m

Now, time taken by ball till first collision with floor :

We know that

s = ut + 1/2 × at^2

=> 80 = 0×t + 1/2 × 10 × t^2

=> 5 × t^2 = 80

=> t^2 = 80/5 = 16

=> t = √16 = 4 sec

NOTE : FIRST PICTURE IS OF EXPLANATION OF THE PROCESS AND SECOND PICTURE IS OF THE GRAPHS.

Explanation is attached with the answer.

Thus, from the above results and calculations, the graph obtained is,

Answer 2

Answer:

Explanation:

distance travelled in 4 sec⇒ 1/2gt^2

                                           ⇒1/2× 10 × 4 × 4

                                           ⇒80 m

distance travelled in 3 sec⇒ 1/2 × 10 × 3 × 3

                                           ⇒45 m

∴ distance travelled in 4 th sec is 80 - 45 =  35 m