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`        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`
3 years ago

```							INITIAL SPEED = uHEIGHT= 80 MTIME OF ASCEND T1:0=u-gT1 TIME OF DESCEND T2:0=u-gT2 SO T1 + T2 =62u/g = 6u = 29.4 m/s  Thanks
```
3 years ago
```							a = g = -10 m/s^(2)s = +80 ms = ut + (1/2) x g x t^(2) ; where t is time and a is acceleration due to gravity80 = ut + (1/2) x -10 x t^(2) 80 = ut – 5{t^(2)}5{t^(2)} – ut + 80 = 0use 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) = 6solving both, you get {[square root](u^2 – 1600)}/5  = 6[square root](u^2 – 1600) =  30u^2 – 1600 = 900u^2 = 2500u = +50 or -50therefore, u = +50 m/s
```
3 years ago
```							a = g = -10 m/s^(2)s = +80 ms = ut + (1/2) x g x t^(2); where t is time and a is acceleration due to gravity80 = ut + (1/2) x -10 x t^(2) 80 = ut – 5{t^(2)}5{t^(2)} – ut + 80 = 0use 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) = 6solving both, you get {[square root](u^2 – 1600)}/5  = 6[square root](u^2 – 1600) =  30u^2 – 1600 = 900u^2 = 2500u = +50 or -50therefore, u = +50 m/s
```
3 years ago
```							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 3sec1-- here u=0 (as velocity is 0 at Max height)Now applying h= ut +1/2 gt^2We get.            h= 0+1/2×10×3×3 =45mNow total height =80+45=125Now 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
```
one year ago
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