my tutor took an hour to explain this. can some1 help?

Question

An astronaut on the planet Zircon tosses a rock horizontally with a speed of 6.35 m/s. The rock falls through a vertical distance of 1.40 m and lands a horizontal distance of 8.55 m from the astronaut. What is the size of the acceleration due to gravity on Zircon?

Answer 1

Well, assuming that Twilo and Zircon have the same gravity...

Newtons first law comes into play here - An object in motion will continue in a straight line at constant velocity unless acted on by another force.

So if the astronaut was in space and threw the rock at 6.35 m/s, it would go on at that speed in a straight line. However, in this case there is a force - from Zircon's gravity - altering the velocity by accelerating the rock downwards. Since velocity is a vector, we can split it out into it's components. The horizontal speed and the vertical speed. Since only the vertical speed is affected by the force from Zircon's gravity, we know that the horizontal speed will not change. This is the key bit to figuring out the rest.

We know that the rock is traveling at 6.35m/s and lands 8.55m away from the astronaut. This tells us how long it takes for the rock to fall the 1.4m vertical distance.

We know that time = distance / speed

So for the rock to travel 8.55m at 6.35m/s takes
(8.55/6.35) = 1.346 seconds.

We now know how long it takes for the rock to fall to the ground.

I'm going to start using some symbols now to reduce the size of the formulae. Nothing to complicated.

Since acceleration is a change in velocity over time, we can figure out the end velocity (we know the starting vertical velocity is 0)

velocity (v) = gravitational acceleration (g) * time (t)

Velocity is also the change in distance over time. If we plotted a graph of velocity (y axis) and time (x axis) then the area under the curve would give us the distance travelled between any two values of x (i.e. time). This is what we do when we perform integration on an equation. So integrating velocity

v = g * t

will give us:

distance (d) = g * (t^2) / 2

Rearranging gives us

g = 2 * d / (t^2)

We know the distance fallen = 1.4m and the time take = 1.346s so running those through the equation gives us:

g = 2 * 1.4 / ( 1.346 * 1.346) = 1.54 m/s^2

This assumes the following:
1 - no air resistance
2 - the throw is perfectly horizontal and the distance from the astronaut is measured from the end of his hand - where he lets go of the rock - to its resting place (we are not told how long the astronaut's arms are so I think that's valid)

Answer 2

This is doable if and only if we assume "horizontal" means equidistant from the surface of the planet. In which case, horizontal is not a straight line, but can be approximated as a straight line if the line is short compared to the circumference of the planet.

t = S/v = 8.55 m/6.35 mps (You can do the math). This is how long it took the rock to fall to the planet and stop S meters away on the horizontal.

D = 1/2 gt^2; where D = 1.40 m vertical drop, g = the acceleration due to gravity you are looking for, and t = the time to drop you found above. Thus,

g = 2D/t^2 (You can do the math)

Lesson Learned: As there are no forces acting horizontally, the horizontal velocity v = constant = 6.35 mps, which will continue until the rock hits the ground from a height D = 1.4 m; so that t = S/v = the time to drop can be found and substituted into D = 1/2 gt^2 to find g.

Answer 3

since there is no external force in horizontal direction,the horizontal speed will remain constant. 6.35 * time =8.55
time of flight =8.55/6.35 =1.34
and during this time only the rock covers vertical distance of 1.40 m.Note that since the rock was thrown horizontally, initial vertical velocity = zero.Assume the acceleration due to gravity on planet is a.
newton's equation: v^2 =u^2 + 2a(s) ........................eq(1)
s = distance travelled
u=0
v = u + at (newtons equation)
v = at
substitue this value of v in eq 1
(at )^2= 0 + 2as
a = 2s / t^2 = 2* 1.40 /(1.34)^2 = 1.55 m/s^2

Answer 4

I think this happens on the planet of Twilo not Zircon.

Answer 5

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Answer 6

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