A coin is thrown in a vertically upward direction with a velocity of $5 m s^{- 1}$ . If the acceleration of the coin during its motion is $10 m s^{- 2}$ in the downward direction, what will be the height attained by the coin and how much time will it take to reach there ?

s = 1.5 m, t = 0.25 s

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s = 0.25 m, t = 1.5 s

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s = 1.25 m, t = 0.75 s

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s = 1.25 m, t = 0.5 s

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Solution

The correct option is D $s = 1.25 m, t = 0.5 s$

Step 1, Given data
Initial velocity, $u = 5 m s^{- 1}$ ,
Acceleration, $a = - 10 m s^{- 2}$ .

Let ' $v$ ' be the final velocity and ' $s$ ' be the height attained.
At the highest point of motion, $v = 0$

Step 2, Finding time

Using third equation of motion, $v^{2} = u^{2} + 2 a s,$
$0 = 5^{2} + [2 \times (- 10) \times s]$
$\Rightarrow s = 1.25 m$
Hence the height attained by the coin is $1.25 m$ .

Let the time taken be $t$ .
Using the first equation of motion,
$v = u + a t$
$0 = 5 - 10 t$
$\Rightarrow t = 0.5 s$
$∴$ Time taken by the coin is $0.5 s$ .

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A stone is thrown in a vertically upward direction with a velocity of

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A coin is thrown in a vertically upward direction with a velocity of 5 ms−1. If the acceleration of the coin during its motion is 10 ms−2 in the downward direction, what will be the height attained by the coin and how much time will it take to reach there ?

A coin is thrown in a vertically upward direction with a velocity of $5 m s^{- 1}$ . If the acceleration of the coin during its motion is $10 m s^{- 2}$ in the downward direction, what will be the height attained by the coin and how much time will it take to reach there ?