 | Samuel Newth - 1864 - 392 σελίδες
...feet per second, to find how high it will rise. By Art. 148, 2gs = V2; therefore in this case, 64-4 Ex. 2. A body is projected vertically upwards with a velocity of 161 feet per second, to find how long it will continue to ascend. y By Art. 149, < = — , therefore... | |
 | Richard Wormell - 1871 - 288 σελίδες
...fall in 5 seconds ? Initial velocity = 30 ft. Final ,, = 30 + 5 x 32 Mean , , Space ,, 550ft. Example 2. A body is projected vertically upwards with a velocity of 100 feet per second ; how far will it ascend in 3 seconds ? Initial velocity = 100 Final „ = 100 —3 x 32 Mean ,, =52... | |
 | Richard Wormell - 1871 - 270 σελίδες
...fall in 5 seconds ? Initial velocity = 30 ft. Final ,, = 30 + 5 x 32 Mean ,, Space ,, 550ft. Example 2. A body is projected vertically upwards with a velocity of 100 feet per second ; how far will it ascend in 3 seconds ? Initial velocity = 100 Final „ = 100 — 3 x 32 Mean „... | |
 | Samuel Newth - 1871 - 152 σελίδες
...will rise. v* By Art. 75. tf = 2<js, or s = - ; therefore, in this case, 2g IOOOO 8=-^ =I55-28. 64-4 Ex. 2. A body is projected vertically upwards with a velocity of 161 feet per second, to find how long it will continue to ascend. By Art. 75, v = gt, or < = -; therefore,... | |
 | Thomas Minchin Goodeve - 1874 - 336 σελίδες
...u- — 2g s = 64^- — 2g x 14 g = 36^. .-. v = ± fig.and 6^= %g — gt, .-. t = 2 seconds. 7T;i-. 2. A body is projected vertically upwards with a velocity...above the point of projection ? Ans. After 2-13 or 4-oS seconds. Ex. 4. A body is thrown vertically upwards with a velocity of 161 feet per second from... | |
 | Stephen Parkinson - 1874 - 436 σελίδες
...as if the particle had moved uniformly during the whole time with the mean velocity. Ex. A particle is projected vertically upwards with a velocity of 100 feet per second, to find (i) its height at the end of 3", and (ii) the time when it is at a height of 140 feet above... | |
 | William Garnett - 1875 - 348 σελίδες
...smooth, is numerically equal to the work done against friction by the former particle. 9. A particle is projected vertically upwards with a velocity of 100 feet per second ; find the time occupied by it in its ascent in describing that portion of its path which lies between the... | |
 | Samuel Newth - 1879 - 370 σελίδες
...find how high it will rise. By Art. 1 78, zgs = V* ; therefore, in this case, i0000 64-4 = i55-28. Ex. 2. A body is projected vertically upwards with a velocity of i6i feet per second, to find how long it will continue to ascend. V By Art. 179, t = — , therefore... | |
 | Thomas Minchin Goodeve - 1883 - 368 σελίδες
...vertically through ng. Here 1f = a2 - 2gs = 64^ - ag x ng = 36^. .-. v = ±6g. and (>g—Zg — gt, .'. t = 2 seconds. Ex. 2. A body is projected vertically upwards...per second : find its height after 3 seconds. Ans. 155-1 feet. Ex. 3. With the same data, after what interval is it 140 feet above the point of projection?... | |
 | Thomas Minchin Goodeve - 1887 - 372 σελίδες
...— zg s = 64^ — zg x 14^ = 36^. . -. z/ = ± 6g. and 6^f = 8 g — gt, .'. t — 2 seconds. j?ir 2. A body is projected vertically upwards with a velocity...per second : find its height after 3 seconds. Ans. 155-1 feet. Ex. 3. With the same data, after what interval is it 140 feet above the point of projection?... | |
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Ex. 2. A body is projected vertically upwards with a velocity of 100 feet per second : find its height after 3 seconds. 
