Question

A body projected vertically up with a speed of 20m/s . The distance travelled by it in 3 sec is (g=10)

Answer 1

Answer:

Explanation:

Time of ascent = u/g

= 20/10

= 2 sec

Maximum height= u²/2g

= 20²/2*10

= 20m

During fall, distance covered in one second,

= 1/2 * 10 * 1²

= 5m

Total distance covered in 3 second

= 20+5

=25 m

Answer 2

Answer:

Explanation:

Time of ascent = u/g

= 20/10

= 2 sec

Maximum height= u²/2g

= 20²/2*10

= 20m

During fall, distance covered in one second,

= 1/2 * 10 * 1²

= 5m

Total distance covered in 3 second

= 20+5

=25 m