Question

Question 11. Ek Ball ko hawa me vertically upwards phenk diya gaya 25 m/s
ki velocity se. Pata kariye ki ye ball hawa me kitni upar jayegi aur kitni der bad
zameen par wapis gir jayegi. (g = 9.8 m/sec)
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Answer 1

Answer:

Hey frnd...

Here is ur ans...

Explanation:

Given u = 25m/s

We know

${v}^{2} - {u}^{2} = - 2gh \\ at \: highest \: pt. \: v = 0 \\ - {25}^{2} = - 2(9.8)h \\ h = \frac{625}{19.6} \\ h= 31.887m$

Also we know

$v \: = \: u - gt \\ \\ 0 = 25 - 9.8t \\ 9.8t = 25 \\ t = \frac{25}{9.8} \\ t = 2.551 \: sec$

Thus half tym of journey = 2.551 sec

Total tym of journey = 2(2.551)

= 5.10204

Hope this helps u

Answer 2

Answer:

We will have to find how much time will the ball take to reach the maximum height.

At maximum height the velocity of the ball will be 0 m/s.

By using v = u+at we can find the time ball will take to reach the max height.

0 = 25 -9.8t

9.8 t = 25

t = 25/9.8 =2.55 sec

Now we will calculate the maximum height the ball will reach.

By using s = ut - (at^2 )/2     [ The acceleration is negative ]

s = 25*2.55 - (9.8 * 2.55*2.55)/2 = 31.89

Now comes the second part of the question in which we have to calculate how much time ball will take to reach the ground.

At the max height velocity of the ball is 0 m/s that is this time we have the initial velocity as 0 m/s

s = ut + ( at^2 )/2

31.89 = (9.8*t^2)/2

63.78 = 9.8 * t^2

t ^2 = 6.5

t = 2.5 sec.