Hi, I have some questions, any help would be appreciated

It's from the Exercise 4.3 of Maths Quest 12 Specialist, if that helps, but the questions are here below.

11 From a hot air balloon rising vertically upward with a speed of 8 m/s, a sandbag is

dropped which hits the ground in 4 seconds. Determine the height of the balloon

when the sandbag was dropped.

12 A missile is projected vertically upward with a speed of 73.5 m/s, and 3 seconds

later a second missile is projected vertically upward from the same point with the

same speed. Find when and where the two missiles collide.

13 A flare, A, is fired vertically upwards with a velocity of 35 m/s from a boat. Four

seconds later, another flare, B, is fired vertically upwards from the same point with

a velocity of 75 m/s. Find when and where the flares collide.

Thanks guys!!!

I haven't done any maths in over two months now so if my solutions are incorrect someone please correct me.

For all of these questions I would use the suvat equations with a=9.8(m/s). In fact, for all of these questions I believe only

is needed.

**11.** In this question you are simply solving for height, or

**s**The question tells you the initial velocity (u) is 8m/s and the time taken to traverse the distance between the position of the hot air balloon when the sandbag is dropped and the ground is 4 seconds, so you now have your u and t values respectively.

Remember that

**the hot air balloon is rising vertically**, and thus you must assign opposite values to u and a (I personally would assign the positive value to the upwards direction).

So your equation would go from:

to

Hence s=-46.4m, and thus the height of the hot air balloon when the sandbag was dropped was 46.4m (in this case the negative sign is indicating that the bag fell in the direction of gravity, ie. downwards).

**12.** This time you have two objects and are trying to find the point at which they collide, thus I would recommend equating their s values.

If we let

**t** be the time passed (in seconds) since the first missile was launched, then the time passed since the second missile was launched would be

**t-3**.

Both missiles have the same initial velocity of 73.5m/s, and both have gravity as their constant acceleration.

Therefore we can equate the values of s for each missile to create the equation

where u=73.5, a=-9.8 and we are solving for t.

Therefore

**t=9 (seconds)** and by substituting that into the original s=ut+1/2at^2 equation we can find s=264.6m

Hence the missiles collide 264.6m above the ground 9 seconds after the launch of the first missile.

**13.** Though the flares don't start with the same initial velocity, we can apply the same methodology in this question as the last. The only difference here would be that the values of u on each side of the equation would differ (we'll use b and c to denote these different initial velocities).

Again, by letting t be the time since the launch of the first flare, the time since the launch of the second flare would be

**t-4**So the equation would go from

to

Thus t=43/9 secs, and by substituting that in s=44849/810m or 55.37m to 2 d.p.

Therefore the flares collide 43/9 (or 4.78 to 2 d.p.) seconds after the launch of the first flare 55.37m above the ground.

Again it's late rn and I haven't done anything remotely mathematical for two months now so if these are incorrect please correct me and I apologise in advance.