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A ball is thrown vertically upwards with an initial velocity of `7sqrt6` ms−1, and is subject to gravity and air resistance. The acceleration of the ball is given by `overset(¨)x = −(9.8 + 0.1v^2)`, where `v` ms−1 is its velocity when it is at a height of `x` metres above ground level.
The maximum height, in metres, reached by the ball is
`A`
`overset(¨)x = v · (dv)/(dx) = −(9.8 + 0.2v^2)`
| `(dv)/(dx)` | `= (−(9.8 + 0.2v^2))/v` |
| `(dx)/(dv)` | `= −(10v)/(98 + v^2)` |
| `x` | `= −int_(7sqrt6)^0 (10v)/(98 + v^2)\ dv` |
| `= 5log_e(4)` |
`=>\ A`
A tourist standing in the basket of a hot air balloon is ascending at 2 ms¯¹. The tourist drops a camera over the side when the balloon is 50 m above the ground.
Neglecting air resistance, the time in seconds, correct to the nearest tenth of a second, taken for the camera to hit the ground is
A. 2.3
B. 2.4
C. 3.0
D. 3.2
E. 3.4
`E`
`u = 2, quad s = -50, quad a = -9.8`
| `2t – 4.9t^2` | `=-50` | |
| `4.9t^2 – 2t – 50` | `= 0` |
`t = (2 +- sqrt((-2)^2 – 4(4.9)(-50)))/(2(4.9))`
`:. t = 3.4\ \ \ (t>0)`
`=> E`