Quick Projectile Question (2)? (1 Viewer)

[Zokunu]

Littlehelp?



StepbyStep would be nice, thank you .

 

[anomalousdecay]

You need to find first when the change in height is zero, when it hits the other cliff, which I am assuming is the same height as the original cliff as fired from.





Now we must find Uy. Draw the triangle and get Uy. First find 92.5 km/hr into m/s, by dividing by 3.6







Now find t from the equation before:





Why does it say t=10 on your sheet?

Also, does it ask for the time of flight for it to fall to the pit, or to hit the cliff?



Also, what is the building?

 

[Zokunu]

You need to find first when the change in height is zero, when it hits the other cliff, which I am assuming is the same height as the original cliff as fired from.





Now we must find Uy. Draw the triangle and get Uy. First find 92.5 km/hr into m/s, by dividing by 3.6







Now find t from the equation before:





Why does it say t=10 on your sheet?

Also, does it ask for the time of flight for it to fall to the pit, or to hit the cliff?



Also, what is the building?

Im pretty sure its meant to be flight for it to fall to the pit, the building is that square thing there are the left and right side. They both have the same height....What do you think?

 

[mysterymarkplz]

I got a different answer to the first guy?
I got 4.98s for the first time of flight, then plus the 10s = 14.98s
92.5km/hr = 24.44m/s
s = ut + 1/2(t^2)
0 = 24.44t - 4.9t^2
t(24.44 - 4.9t) = 0
t = 24.44/4.9
= 4.98
Or you can do it with v = u + at
0 = 24.44 - 9.8t
t = 24.44/9.8
= 2.49s
since that is t when v = 0, times t by 2 you get 4.98s, same as the s=ut+(1/2)at^2 equation.

And for the second part of the question,
Since its a cliff scenario, where there is no u(y) and only ux,
s = ut + (1/2)at^2
s = 0 -4.9(10^2)
s = -490m
490m.

 

[panda15]

Hmm. This question is pretty ambiguous, but here's how I would go about it:

 

[Zokunu]

Hmm. This question is pretty ambiguous, but here's how I would go about it:

mhm, i agree with you. Mystery, i THINK you forgot about the ut = 24.43 for the second question. Can someone please confirm if this answer is correct?

Anyways, thank you all .

 

[panda15]

mhm, i agree with you. Mystery, i THINK you forgot about the ut = 24.43 for the second question. Can someone please confirm if this answer is correct?

Anyways, thank you all .

It's there, I just skipped a couple steps. In the first part, I just divided everything by t, giving 0=24.43-4.9t
In the second part, t=10 and u=-24.43, hence ut equaling -244.3.

 

[mysterymarkplz]

I thought ut = 0? Its like those scenarios where a plane drops a bomb, the bomb has 0 u(y) but has a value for u(x), similarly in your question the ball hits the wall, its vertical velocity is only influenced by gravity? o.o

 

[panda15]

I thought ut = 0? Its like those scenarios where a plane drops a bomb, the bomb has 0 u(y) but has a value for u(x), similarly in your question the ball hits the wall, its vertical velocity is only influenced by gravity? o.o

But at the point where it hits the wall, the projectile has vertical velocity from the initial projection. Given that it hits the building at the same height that it is fired from, it has the same magnitude of velocity as it does initially at the point where it hits the building, hence u=-24.43.