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A ball is thrown upward and caught by the thrower after 8 seconds.How below the highest point was it 5 seconds from the start? 1m 4.9m 9.8m 19.6m PLS ANSWER FAST
A ball is thrown upward and caught by the thrower after 8 seconds.How below the highest point was it 5 seconds from the start?	1m	4.9m	9.8m	19.6mPLS ANSWER FAST

```
3 years ago

Arun
25768 Points
```							8 second here indicates the total time taken by the ball to reach highest point and come back to the thrower. So the highest point is reached at half the time i.e. 4 seconds. take g=9.8m/s^2v=u-gt (since ball goes upward)at highest point v=0, time taken=4u=v+gtu=39.2m/s…….v^2=u^2-2gh (ball goes upward)at highest point v=0h=u^2/2gh=78.4m……..(highest point)h=ut- .5gt^2t=5 seconds, u=39.2h=196 - 0.5 *9.8 *25h = 73.5so your answer would be= 78.4–73.5= 4.9m
```
3 years ago
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