I NEED major help, its about getting time?

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To solve the problem of a stone thrown horizontally from a cliff, the time of flight can be calculated using the vertical motion equation, where the height of the cliff is 78.4 m and the acceleration due to gravity is 9.8 m/s². The horizontal distance traveled can then be determined using the horizontal velocity of 8.0 m/s and the time calculated from the vertical motion. The GUESS method is mentioned but not clearly defined, with participants suggesting that it involves breaking down the motion into horizontal and vertical components. A similar approach is applied to a dart throw problem, emphasizing the independence of x and y motions and the need to find the common variable, which is time. Overall, the discussions focus on applying basic physics equations to solve projectile motion problems.
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How would you get time for this:

5. A stone is thrown horizontally at 8.0 m/s from a cliff 78.4 m high. How far from the base of the cliff does the stone strike the ground? (Use GUESS Method)

G: vi = 8.0 m/s
g = 9.8 m/s2
dy = 78.4 m
U: dx
E: d = vi*t +1/2 a*t^2
S: dx = 32 m


vi =initial velocity
^2= squared
*= times
 
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The motion in the y-direction and in the x-direction are independant, right?

So first you'll want to use equation "E" to calculate the time is takes the rock to fall the 78.4m. Then, use the time you found in a similar equation, but for the x distance (of course is no acceleration term in equation "E")
 
I don't know what the GUESS method is, but the first thing I would do is ignore that any type of horizontal motion exists. I would imagine the stone simply being dropped off the cliff. With the knowledge of the height of the cliff and the fact that it is dropped (vi = 0) you can calculate the amount of time that the stone is in the air. Now I remind myself of this horizontal velocity that the rock was given. To calculate the horizontal distance I just use d = vi*t +1/2 a*t^2. There is no acceleration in the x direction so it is simply D = ViT. You know the initial velocity and you calculated the time.
 
...But how would i get the time, i don't know how to get it?
 
feod2003 said:
How would you get time for this:

5. A stone is thrown horizontally at 8.0 m/s from a cliff 78.4 m high. How far from the base of the cliff does the stone strike the ground? (Use GUESS Method)

G: vi = 8.0 m/s
g = 9.8 m/s2
dy = 78.4 m
U: dx
E: d = vi*t +1/2 a*t^2
S: dx = 32 m


vi =initial velocity
^2= squared
*= times
I don't know what the GUESS method is. However, you should note that v_i is 8m/s in the x direction. V_xi =8. There is no initial component of v in the y direction. V_yi = 0. In the x direction, there is no acceleration, so v_x stays constant throughout the trip, therfore, x =v_x(t), this is the horizontal distance traveled. To solve for t, use your the "E" equation in the y direction, where d =78.4 , v_i = 0, and a =9.8. Then solve for x, I guess.
 
PhanthomJay said:
I don't know what the GUESS method is. However, you should note that v_i is 8m/s in the x direction. V_xi =8. There is no initial component of v in the y direction. V_yi = 0. In the x direction, there is no acceleration, so v_x stays constant throughout the trip, therfore, x =v_x(t), this is the horizontal distance traveled. To solve for t, use your the "E" equation in the y direction, where d =78.4 , v_i = 0, and a =9.8. Then solve for x, I guess.

wow thanks that helped a bit, thanks guys
 
I got another one.

7.A dart player, standing 3.2 m away from the dartboard, throws a dart horizontally at a speed of 12.4 m/s. How far from the bull’s eye (where the player aimed) did the dart strike? (Use GUESS Method)

G: g = 9.8 m/s2
dx = 3.2 m
vi = 12.4 m/s
U: dy
E: d = vit +1/2at2
S: dy = 0.32 m (give a direction)
 
need more help fast.?

7.A dart player, standing 3.2 m away from the dartboard, throws a dart horizontally at a speed of 12.4 m/s. How far from the bull’s eye (where the player aimed) did the dart strike? (Use GUESS Method)
 
I think people on these boards are more familiar with teh ViVfdat method (same sort of thing...)

again same sort of concept as ur last problem...no initial y velocity...no x acceleration...then just set up a system with the values...knowing that acceleration in y direction is g and then solve...its similar to the last one u did if you got it...just having a different unknown this time...meanign a different equation...

Vix 12.4
Vfx 12.4
Dx 3.2
ax 0
t = t

Viy = 0
Vfy = ?
dy = x
ax = -g
t = t

whats the only common variable between the two sets?
 
  • #10
Mehta29 said:
I think people on these boards are more familiar with teh ViVfdat method (same sort of thing...)

again same sort of concept as ur last problem...no initial y velocity...no x acceleration...then just set up a system with the values...knowing that acceleration in y direction is g and then solve...its similar to the last one u did if you got it...just having a different unknown this time...meanign a different equation...

Vix 12.4
Vfx 12.4
Dx 3.2
ax 0
t = t

Viy = 0
Vfy = ?
dy = x
ax = -g
t = t

whats the only common variable between the two sets?

umm... is that guna be t?
 
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