Question

A driver drops from a 10min high chiff. At what speed does he enter in the water and how long he is in the air?​

Answer 1

Answer:

Applying 3rd equation to calculate the final velocity,

v

2

=0+2∗9.8∗40

v =

784

= 28 m/s .

Time taken, t

1

=

9.8

28

[ u + at ]

= 2.85 s.

Then, the second part says that he reached the ground with

velocity 2 m/s (final velocity) with deceleration of 2 m/s

2

and the initial velocity would be which we calculated above i.e. 28m/s

Assume, it traveled a distance of 'h' while falling.

Again, apply 3rd equation of motion,

(2)

2

=(28)

2

+2(−2)∗h

4 * v = 784 - 4

= 780

h = 195m.

Time taken, t

2

=

2

28−2

[ v = u + at ]

= 13 s

Time taken while he is in air,

t = t

1

+t

2

= 2.85 + 13

= 15.85 s

Explanation:

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