Question

whats its answer? i cant find it and please explain it , its of 10 std science 2 chapter 1​

Answer 1

Sign convention : Consider + ve sign For upward movement and - ve sign for downward mivement

Height , H = displacement = s = + 4.05 m

( as ball displaces from lower level to upper )

Let u = initial velocity

When ball reaches to height h , the final velocity v become 0 . so , v = 0 m / s

Here g = acceleration due to gravity ( acting downward towards the Earth ) = - 10 m / s^2

Now using 3rd equation of motion , we get

v^2 = u^2 + 2 × g × s

=> u^2 = v^2 - 2 × g × s

= 0 ^2 - 2 × ( - 10 ) × 4.05

= 0 - ( - 81 ) = 81

=> u^2 = 81 => u = root ( 81 ) = + - 9 m / s

=> u = + 9 m/s

( as ball is moving from lower to higher level )

So , the initial speed with which ball is thrown

u = 9 m /s

Let t = time taken by ball to reach the height h

Niw Using first equation of motion

v = u + a × t = u + g × t ( a = g )

=> t = ( v - u ) / g = ( 0 - 9 ) / ( - 10 ) = 9 / 10

=> t = 0.1 s

Total time taken by the ball in its whole journey

= 2 × t = 2 × 0.1 = 0.2 s

Answer 2

Explanation:

Given : Height (h) =4.05 m

g =10m/s^2

To find : Initial velocity (u) =?

Time Taken (t) =?

Solution :

When ball reaches to a height of 4.05m, it comes to rest for some instant of time and then return to back

Hence' Final velocity (v) at that moment will zero

V = 0

V ^2 = u^2 - 2gh

0 = u^2 - 2×10×4.05

u^2 =81

u = 9m/s ans.

Time Taken to reach the height 4.05m

h = ut - 1/2gt^2

4.05= 9t - 5t^2

5t^2 - 9t + 4.05 =0

$t = \frac{ 9 + - \sqrt{ ({ - 9)}^{2} - 4 \times 5 \times 4.05 } }{ 2 \times 5} \\ = \frac{9 + - \sqrt{81 - 81} }{2 \times 5} \\ = \frac{9}{10} \\ = .9 \: sec$

Total time taken = time taken to go upward + to go downward

=. 9+.9

t =1.8 sec