Aussie Maths & Science Teachers: Save your time with SmarterEd
From an open window, a person projects a ball vertically up using an outstretched arm so the ball does not strike any part of the building. The point of projection of the ball is 50 m above the ground and its velocity of projection is 20 m s\(^{-1}\).
The time, in seconds, it takes for the ball to reach the tray of a truck that is 1 m above the ground directly below the point of projection is closest to
\(B\)
\(u=20\ \text{ms}^{-1}, \ a=-9.8\ \text{ms}^{-2}, \ s=-49\)
\(\text{Find}\ t\ \text{when}\ \ s=-49:\)
\(\text{Using}\ \ s=u t+\dfrac{1}{2} a t^2,\)
\(-49=20 t-4.9 t^2 \ \ \Rightarrow \ \ t=-1.72,5.80\)
\(\Rightarrow B\)
A tourist in a hot air balloon, which is rising vertically at 2.5 m s\(^{-1}\), accidentally drops a phone over the side when the phone is 80 metres above the ground.
Assuming air resistance is negligible, how long in seconds, correct to two decimal places, does it take for the phone to hit the ground?
\(E\)
\(\text{Take upward velocity as positive.}\)
\(u=2.5\ \text{ms}^{-1}, \ a=-9.8\ \text{ms}^{-2} \)
\(\text{Using}\ \ s=ut+\dfrac{1}{2}at^2,\)
\(\text{Find}\ t\ \text{when}\ \ s=-80\ \text{(by calc):} \)
\(-80=2.5t-4.9t^2\ \ \Rightarrow \ \ t=4.30\ \text{s}\)
\(\Rightarrow E\)
At her favourite fun park, Katherine’s first activity is to slide down a 10 m long straight slide. She starts from rest at the top and accelerates at a constant rate, until she reaches the end of the slide with a velocity of `6\ text(ms)^(-1)`.
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When at the top of the slide, which is 6 m above the ground, Katherine throws a chocolate vertically upwards. The chocolate travels up and then descends past the top of the slide to land on the ground below. Assume that the chocolate is subject only to gravitational acceleration and that air resistance is negligible.
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At what velocity would the chocolate need to be propelled upwards, if Katherine were to immediately slide down the slide and run to reach the chocolate just as it hits the ground?
Give your answer in `text(ms)^(-1)`, correct to one decimal place. (2 marks)
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Katherine’s next activity is to ride a mini speedboat. To stop at the correct boat dock, she needs to stop the engine and allow the boat to be slowed by air and water resistance.
At time `t` seconds after the engine has been stopped, the acceleration of the boat, `a\ text(ms)^(-2)`, is related to its velocity, `v\ text(ms)^(-1)`, by
`a = -1/10 sqrt(196-v^2)`.
Katherine stops the engine when the speedboat is travelling at `7\ text(m/s)`.
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a. `u = 0, quad v = 6, quad s = 10`COMMENT: Exact form required. 3.3 seconds was marked incorrect!
| `text(Solve for)\ \ t:\ \ \ ((u + v)/2)t` | `= s` |
| `((0 + 6)/2)t` | `= 10` |
| `t` | `= 10/3\ text(s)` |
b. `u = 10, quad a = -9.8, quad s = -6`COMMENT: Many students showed a “lack of understanding” of displacement here.
`text(Solve for)\ \ t:`
| `s` | `= ut + 1/2 at^2` |
| `-6` | `= 10t-4.9 t^2` |
| `:. t` | `~~2.5\ text(s)\ \ \ text{(by CAS)}` |
♦♦ Mean mark 27%.
c. `text(Time of chocolate in air)`
`= 10/3 + 4`
`=22/3`
`text(Solve for)\ \ u:`
| `-6` | `= 22/3 u -4.9(22/3)^2` |
| `:. u` | `~~35.1\ text(m s)^(−1)\ \ \ text{(by CAS)}` |
| d.i. | `(dv)/(dt)` | `= -1/10 sqrt(196-v^2)` |
| `(dt)/(dv)` | `= (-10)/sqrt(196-v^2)` | |
| `t` | `= int(-10)/sqrt(196-v^2)\ dv` | |
| `-t/10` | `= int 1/sqrt(14^2-v^2) dv` | |
| `-t/10` | `= sin^(-1)(v/14) + c` |
`text(When)\ \ t = 0, v = 7`
`=> c=-sin^(-1)(1/2) = -pi/6`
| `-t/10` | `=sin^(-1)(v/14)-pi/6` | |
| `sin^(-1) (v/14)` | `= pi/6-t/10` | |
| `:. v` | `=14sin(pi/6-t/10)` |
d.ii. `text(Find)\ \ t\ \ text(when)\ \ v=0:`
| `-t/10` | `=sin^(-1)(0)-pi/6` | |
| `t` | `= -10 sin^(-1)(0) + (10 pi)/6` | |
| `= (5 pi)/3\ text(s)` |
d.iii. `v = 14sin(pi/6-t/10)`
| `(dx)/(dt)` | `=14sin(pi/6-t/10)` |
| `x` | `= int_0^((5pi)/3)14sin(pi/6-t/10)\ dt` |
| `~~ 18.8\ text(m)\ \ \ text{(by CAS)}` |
A tourist in a hot air balloon, which is rising at 2 m/s, accidentally drops a camera over the side and it falls 100 m to the ground.
Neglecting the effect of air resistance on the camera, the time taken for the camera to hit the ground, correct to the nearest tenth of a second, is
A. 4.3 s
B. 4.5 s
C. 4.7 s
D. 4.9 s
E. 5.0 s
`C`
`u = 2, s = −100, a = -9.8`
| `s` | `= ut + 1/2at^2` |
| `-100` | `= 2t – 4.9t^2` |
| `0` | `= 4.9t^2 – 2t – 100` |
`text(Solve:)\ \ 4.9t^2 – 2t – 100=0\ \ text(for)\ \ t`
`t~~4.7\ \ text(sec)`
`=> C`
A ball is thrown vertically up with an initial velocity of `7sqrt6` ms`\ ^(–1)`, and is subject to gravity and air resistance.
The acceleration of the ball is given by `ddotx = −(9.8 + 0.1v^2)`, where `x` metres is its vertical displacement, and `v` ms`\ ^(–1)` is its velocity at time `t` seconds.
The time taken for the ball to reach its maximum height is
A. `pi/3`
B. `(5pi)/(21sqrt2)`
C. `log_e(4)`
D. `(10pi)/(21sqrt2)`
E. `10log_e(4)`
`D`
`(dv)/(dt) = −(9.8 + 0.1v^2)`♦ Mean mark 42%.
`(dt)/(dv) = −1/(9.8 + 0.1v^2)`
`text(Max height occurs when)\ \ v=0.`
`text(Find)\ \ t\ \ text(when)\ \ v=0:`
| `t` | `= int_(7sqrt6)^0 −1/(9.8 + v^2/10)\ dv` |
| `= (10pi)/(21sqrt2)` |
`=> D`