Question

An object is thrown vertically upwards and rises to a height of 10 m. Calculate (1) the velocity with wich the object was thrown upwards and (2) the time taken by the object to reach the highest point ​

Answer 1

$h \: = 10m$

$g = 10m/s {}^{2} ↧$

$u = ?$

using :

$2as = v {}^{2} - u {}^{2}$

Therefore,

$⤷2(10)(10) = (0) {}^{2} - (u) {}^{2}$

$⤷u {}^{2} = 200$

$⤷u = 10 \sqrt{2} m/s$

therefore,

$⤷v \: = u + at$

$⤷0 = 10 \sqrt{2} - 10t$

$⤷ \: t \: = \sqrt{2} sec.$

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Answer 2

Answer:

The velocity with which the object was thrown upwards = 14 m/s

The time taken by the object to reach the highest point = 1.43 s

Explanation:

An object is thrown vertically upwards. The maximum point travelled by it is the distance of the object. It is given as, s = 10 m.

The final velocity, v of the object is 0. Because the object comes to rest at maximum height and then falls.

Since the motion of the ball is upwards the value of acceleration due to gravity is taken as negative.

Initial velocity, u can be found using the 3rd equation of motion.

   v² = u² + 2as

⇒ u² = v² - 2as

        = 0² - (2 × -9.8 × 10)

        = - (-196)

∴   u = √196 = 14 m/s

Now the time taken can be found using the 1st equation of motion.

  v = u + at

⇒ t = (v - u) / a

     = (0 - 14) / (-9.8)

     = -14 / -9.8

∴ t = 1.43 s