Question

a boy drops a water balloon from the window of a car moving at 104.4 km/h. If the balloon drops from a height of 1.56 m, what horizontal distance will the balloon travel from the release point to where the balloon hits the ground?

Answer 1

first convert Km/h to m/s to find v

consider the vertical component of the motion
s = ut - 16t^2

where s = distance, u = initial vertical velocity and t = time

s = 1.56, u = 0

so 1.56 = 16t^2

find t

t is the time the balloon is in the air

Consider horizontal movement:

distance travelled in meters = velocity * time = v * t

Answer 2

* correction in third line

s = ut + 16t^2

Answer 3

another correction !!

the 16 is not correct - thats the constant for acceleration in feet and we are dealing with meters here.

the formula should s = ut + 4.9t^2

Answer 4

Nice question. First we must determine the total time for travel, so I will use the following kinetics question:\[\Delta(y)=v _{o}+\frac{ 1 }{ 2 }(g \times t ^{2})\]We know that y = 1.56m, and that g = 9.8m/s^2 (I'm rounding to 10 for ease of calculation). So, plus in the values, and t should be approximately 0.54772s (remember, g was rounded to 10).

Next step is converting the velocity 104.4 km/hr to m/s, you should get 29 m/s. If you are unsure, let me know....I don't mind at all.

Then, we use another kinetics equation:\[\Delta(x)=vt=\left( \frac{ v _{o}+v }{ 2 } \right)t\]
Which then is \[29 m/s \times 0.54772s\] = 15.8m or 16m. Hope this helps!