Question

a stone is vertically thrown upwards went up 98m and came down. how long was it in the air?

Answer 1

$\large \underline{ \pmb {\orange{{ \frak{Given : - }}}}}$

• Maximum height = 98m
• Final velocity = 0m/s

final velocity is assumed to be 0 because after reaching maximum height it comes to rest momentarily.

$\large \underline{ \pmb{\orange{{ \frak{Required \: Answer : - }} }}}$

By using 3rd kinematical equation we will find initial velocity :-

• $\underline{\boxed{\frak{{v}^{2} - {u}^{2} = 2as }}}\bigstar$

→ $\sf {(0)}^{2} - {u}^{2}= 2(-9.8)(98)$

→ $\sf {u}^{2} = 1920.8$

→ $\sf u = 43.9 m/s$

Now, by using 1st kinematical equation we will find time taken to reach maximum height :-

• $\underline{\boxed{\frak{v = u+at }}}\bigstar$

→ $\sf 0 = 43.9 + (-9.8)t$

→ $\sf 9.8t = 43.9$

→ $\sf t = 4.4s$

• Now, the time taken to reach the maximum height will be equal to the time taken to return from the maximum height.

• hence, in order to calculate the time of flight we will simply multiply it with 2 in order to get the required answer

• time of flight = 2×4.4 = 8.8 seconds.

• Your answer will be approximately equal to 9 seconds.

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Hope it's beneficial :D