Question

A stone thrown vertically upwards went up 98 m and came down. How long was it in the air?

Answer 1

Answer:

From the 3rd equation of motion, we know that v^2 - u^2 = 2gh. When the stone is going up, we have, v=0, g=-9.8m/s^2, and h=98m.

Crunching the numbers we get, u=43.826 m/s.

Now using the 1st equation of motion,

v=u+gt, again v=0 and g=-9.8m/s^2.

We can use this equation to calculate the time it takes to reach 98m.

Using the equation, time comes out to be 4.47 seconds. The stone will take another 4.47 seconds to come back.

Therefore, total time it is in the air=8.94 seconds

Explanation:

please follow me

mark as brainliest

Answer 2

Hey mate here is your answer

Given :-

Hmax = 98 m

To find :-

Time it was in the air

Formulae to be used :-

1) Hmax = u²/2g

2) Time of flight = 2u/g (Time of flight is the time for which the ball remained in air)

How to find :-

Hmax = u²/2g = 98

Hence u² = (98)*(19.6)         (g = 9.8 m/s²)

u² = 1,920.8

u = 43.82 m/s

Now ,

Time of flight = 2u/g = 2*(43.82) / 9.8

Time of flight = 87.64 / 9.8 ≈ 9 s

Hence the ball remained in air for 9 s.

HOPE IT HELPS YOU

PLEASE MARK ME AS THE BRAINLIEST