Question

This question is really throwing me off.
Michael throws a rock on Planet X at an angle of 59.4◦ with a speed of 30.2 m/s. It
reaches a maximum height of 215 m before falling back down to the ground.

(a) What is the acceleration due to gravity on Planet X, gX? Write your answer
first in units of m/s2 and then in units of g.
(b) How far along the x-direction does the rock travel?

Answer 1

How would you solve for part b

Answer 2

once you have \(g_X\) from part a .... you can plug-it into an equation to solve for time \((\Delta t)\)....

then you can solve for \((\Delta x)\)

Answer 3

is your initial velocity in this question 30.2 m/s? and final velocity 0 m/s?

Answer 4

for part a)...??

Answer 5

yes

Answer 6

and i guess for part b as well

Answer 7

for part a)
http://openstudy.com/users/paxpolaris#/updates/4fcea5c6e4b0c6963ad9ae0a

Answer 8

okay so would u use t=v/a to solve for t, then d=vot + 0.5at^2

Answer 9

and for this part b) is vo = 30.2

Answer 10

well sorry 30.2sin59.4

Answer 11

you could use either t= v/a ...
OR \(\large\Delta y=v _{oy} \Delta t + \frac12 a _{y} \Delta t ^{2}\) to find time \[\Large v _{oy} = 30.2\sin(59.4\deg)\]

if you use the first, or if you use d= 215 ... then you need to double the time

Answer 12

if you use d=Δy= 0 ... then you don't need to double the time

Answer 13

to answer b) you have to finally use:
\[\large \Delta x = v _{ox} \Delta t\]

where: \(\huge v _{ox} = 30.3\cos \left( 59.4\deg \right)\)

Answer 14

*** 30.2 cos 59.4

Answer 15

sorry i dont understand what youre saying with all those symbols

Answer 16

we have to deal with the x component and y component separately..

for the y (vertical) component ... we have constant acceleration \((-g_X)\).
We can use the formulas: \[\large v=u+at\]\[\large s= ut+\frac12at^2\]\[\large v^2=u^2+2as\]for the y components....where u=initial velocity, s=displacement.

Answer 17

for the x (horizontal) component... we have 0 acceleration ... we have constant velocity.

so, in the second equation above ... if you plug in a=0 ... you get for the x-component
\[s=ut\]

Answer 18

to separate the initial velocity to its x\y components...

\[\large u_x=v_{ox}= 30.2\cos(59.4)\]\[\large u_y=v_{oy}= 30.2\sin(59.4)\]