A ball is thrown upward with initial velocity u at a height of 80m at 2 times The time interval being 6 sec find u

INITIAL SPEED = u

HEIGHT= 80 M

TIME OF ASCEND T1:

0=u-gT₁

 

TIME OF DESCEND T₂:

0=u-gT₂

 

SO T_{1 +} T₂ =6

2u/g = 6

u = 29.4 m/s 

 

Thanks

a = g = -10 m/s^(2)
s = +80 m

s = ut + (1/2) x g x t^(2) ; where t is time and a is acceleration due to gravity

80 = ut + (1/2) x -10 x t^(2)
 
80 = ut – 5{t^(2)}

5{t^(2)} – ut + 80 = 0

use the quadratic equation to solve for ‘t’

t = {u + [square root] (u^2 – 1600) }/10 --- eq.(i)    &   {u – [square root](u^2 – 1600)}/10 --- eq.(ii)

eq.(i) + eq.(ii) = 6

solving both, you get 

{[square root](u^2 – 1600)}/5  = 6

[square root](u^2 – 1600) =  30

u^2 – 1600 = 900

u^2 = 2500

u = +50 or -50

therefore, u = +50 m/s 

 

a = g = -10 m/s^(2)
s = +80 m

s = ut + (1/2) x g x t^(2); where t is time and a is acceleration due to gravity

80 = ut + (1/2) x -10 x t^(2)
 
80 = ut – 5{t^(2)}

5{t^(2)} – ut + 80 = 0

use the quadratic equation to solve for ‘t’

t = {u + [square root] (u^2 – 1600) }/10 --- eq.(i)    &   {u – [square root](u^2 – 1600)}/10 --- eq.(ii)

eq.(i) + eq.(ii) = 6

solving both, you get 

{[square root](u^2 – 1600)}/5  = 6

[square root](u^2 – 1600) =  30

u^2 – 1600 = 900

u^2 = 2500

u = +50 or -50

therefore, u = +50 m/s 

 

In the time given (6sec) it first reaches a Max height in 3 sec after the height 80m and then again takes 3 sec to to come back to the point of 80 m(which means the time 6sec is broken in 2 parts of reaching a Max height and returning back)

 

Now first we will find the Max height it attains after 80m in 3sec

1-- here u=0 (as velocity is 0 at Max height)

Now applying h= ut +1/2 gt^2

We get.            h= 0+1/2×10×3×3 =45m

Now total height =80+45=125

Now applying v^2=u^2 - 2gh to final condition we get.        v^2+2gh = u^2

               0+2×10×125 = u^2.  (v=0 coz at h(max) velocity is 0 and we are considering the whole height)

                    u = 50 m/s