Question

Please please solve question 11

Answer 1

for 1 case
h = 1/2gt^2
t = √45*2/10 = √9.0 = 3
now , for2 case
h = ut + 1/2gt^2
45 = u*(3-1) + 1/2*10*(3-1)^2
u = 45 - 5*(3)^2/3
u = 45 - 15/3
u = 12.5 .

Answer 2

$\pink{\underline{\huge{Solution}}}$

When a body is reached to the ground from the height 'h' then the initial velocity (u) of body becomes zero.
[vice versa]

In case 1 velocity (initial) of
body becomes zero ( and both balls reach ground at same time.So, we taking a condition first and time be 't sec ')

$\red{\underline{PART-1}}$

using second formulae of motion

$S = ut+\frac{1}{2}\times at^{2}$

u = 0 m/sec
t = ?
a = g = 10 m/sec
S = 45 m

$S = ut+\frac{1}{2}\times at^{2}\\45 = 0(t)+\frac{1}{2}\times 10(t)^{2}\\45 =\frac{1}{\cancel 2}\times \cancel{10} t^{2}\\5t^{2} = 45 \\ t^{2} = \frac{\cancel 45}{\cancel 5}\\t =\sqrt 9 \\ t = 3 sec$

Now we get ball takes ''3 sec'' to reach the ground and in question we have given that the second ball dropped after 1 sec of second ball then it's time is (3-1) sec = (2 sec)


$\red{\underline{PART-2}}$

S = 45 m
t = (3s-1s) = 2s
a = g = 10 m/sec
u = ?

$S = ut+\frac{1}{2}\times at^{2}\\45 = u(2)+\frac{1}{2}\times 10(2)^{2}\\45 =2u + \frac{1}{\cancel 2}\times \cancel{10}\times 4\\ 45 = 2u +20\\ 2u = 45-20 \\ u =\frac{25}{2} \\u = 12.5 m \:sec^{-1}$

$\orange{\boxed{\bold{u=12.5\:m\:sec^{-1}}}}$