How Do You Calculate Projectile Motion for a Stone Thrown Upward from a Cliff?

[Jenybeny27]

Could someone please try and help me figure out this problem? I keep getting very large numbers and I do not think it is correct! Thanks

A stone is thrown with an initial velocity of 20 m/s straight upward from the edge of a cliff 100m above a canyon floor. The stone just misses the eliff's edge on its way down.

1.) determine the time required for the stone to reach its maximum height.
2.) determine the maximum height of the stone above the edge of the cliff.
3.) how much time elapses as the stone falls from its maximum height to the level from which it was thrown?
4.) what is the velocity of the stone upon returning to the level from which it was thrown?
5.) Determine the velocity of the stone 6 seconds after it is thrown.
6.) determine the position of the stone 6 seconds after it is thrown.

Sorry its long, but pelase help!

 

[Doc Al]

Show us your work so we can check it out.

 

[Jenybeny27]

okay!

1.) Vi= 20m/s (given)
Vf= 0m/s (because @ max height, Vf is 0)
a = 9.81m/s^2

a= change in v/t
9.81m/s2=20m/s / t
t = 2.04 seconds (I think i did that right.)

2.) v=20m/s t=2.04s d=vt d=(20m/s)(2.04s) d= 40.8 m

((i have more work, i just want to make sure that's correct before i proceed to type it all))

 

[Jenybeny27]

3.) how much time elapses as the stone falls from its maximum height to the level from which it was thrown?
a=9.81m/s2 d= 40.8m d=(1/2)at^2 40m=(1/2)9.81m/s2(t^2) t= 2.85s

 

[Doc Al]

1.) Vi= 20m/s (given)
Vf= 0m/s (because @ max height, Vf is 0)
a = 9.81m/s^2

a= change in v/t
9.81m/s2=20m/s / t
t = 2.04 seconds (I think i did that right.)

Right!

2.) v=20m/s t=2.04s d=vt d=(20m/s)(2.04s) d= 40.8 m

Incorrect. The equation d = vt assumes the speed is constant, but that's not true here. The motion is uniformly accelerated. What are some useful kinematic equations for uniformly accelerated motion?

 

[Doc Al]

3.) how much time elapses as the stone falls from its maximum height to the level from which it was thrown?
a=9.81m/s2 d= 40.8m d=(1/2)at^2 40m=(1/2)9.81m/s2(t^2) t= 2.85s

Incorrect, because you are using the incorrect distance. (Also realize that the time it takes for the stone to rise to maximum height must equal the time it takes to fall back down to that same starting point.)

 

[Jenybeny27]

What are some useful kinematic equations for uniformly accelerated motion?

Um, the equations i ahve are:
a = v/t
Vf= Vi+at
d=Vit +(1/2)at^2
Vf^2= Vi^2 + 2ad

I think I would use the d= vit + (1/2)at^2 equation, right?

Therefore, d= (20m/s)(2.04s) + (1/2)(9.81m/s^2)(2.04s)^2
d= 40.8 + 20.4
d= 61.2m (maybe?)

 

[Doc Al]

Um, the equations i ahve are:
a = v/t
Vf= Vi+at
d=Vit +(1/2)at^2
Vf^2= Vi^2 + 2ad

All good.

I think I would use the d= vit + (1/2)at^2 equation, right?

Right!

Therefore, d= (20m/s)(2.04s) + (1/2)(9.81m/s^2)(2.04s)^2
d= 40.8 + 20.4
d= 61.2m (maybe?)

Careful about directions (which are represented by signs). If you take up as positive, then the initial velocity is +20 m/s (since it's upward) but the acceleration is -9.81 m/s^2 (since it acts down). Do it over with the correct sign for a.