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- A ball is thrown and projected upward with angle of elevation with respect to the horizontal. The ball reaches its maximum height in 1.5 seconds and covers a horizontal distance of 200 ft. Calculate the initial velocity of the and the angle of elevation it makes with the horizontal.
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Let the angle of projection with the horizontal be `theta` .
Then the vertical component of velocity is `usintheta` , and the horizontal component of velocity is `ucostheta` , where u is the initial velocity of projection.
It reaches a maximum height in 1.5 seconds.
Analyzing the vertical component, at the maximum height
`rArr usintheta=1.5*32.2` --- (i)
Again considering the horizontal component, horizontal distance travelled by this time is `ucostheta*1.5` .
This distance is half its horizontal range.
`rArr ucostheta=100/1.5=200/3` --- (ii)
Therefore, the initial velocity of projection was 82.3 ft/sec. and angle of projection was `36^o` .
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