Solving a Projectile Motion Problem

[wuffle]

Hello! Thank you forum for providing help for my last thread :).

So far, have 2 more problems which I have some idea how to solve, but not enough to solve it.

Homework Statement

A stone is projected at an initial speed of 120 ft/s directed 62 degrees above the horizontal , at a cliff of height h. The stone strikes the ground 5.5 s after launching. Find the height of the cliff, the speed of the stone just before impact at A and maximum height H reached above the ground

Pretty much, its ball thrown from ground and landing at a cliff.

Homework Equations

The Attempt at a Solution

I think I did it, but I am 100 % sure I got something wrong, here's what I did:

First I found V0y=v0 * sin 62 =32.3 m/s

Next I found the time, when the ball has reached its maximum height.
t=(v0y-Vy)/g=(32.3-0)/9.8=3.3 s
Found the max height
ymax=v0yt-(gt^2)/2=32.3 * 3.3 -(*0.5*9.8 * 3.3^2)= 53.229 m.

Now this is where I think I got it wrong.

Now we calculate the speed of the stone JUST BEFORE THE IMPACT at A.

We know that the time it took the stone to get from the maximum height to cliff is =5.3-3.3 = 2 s

V=-gt(because the velocity at the maximum height of the stone is 0)
V=-21.56 m/s
as we need speed, its absolute value aka =21.56 m/s

We found the speed of the stone, max height H, let's find how many meters did the stone travel from the maximum height to the cliff.

S(from max height to cliff)=gt^2 *0.5=4*0.5*9.8 = 19.6 m(cuz velocity at the maximum height is 0)

So that's it, answers are
S(from max height to cliff) =53.229-19.6=33.629 m
Speed is 21.56 m/s
S(max)=53.229 m

Where did I made a mistake? I know I did it somewhere because I missed something, help please!

 

[PeterO]

Hello! Thank you forum for providing help for my last thread :).

So far, have 2 more problems which I have some idea how to solve, but not enough to solve it.

Homework Statement

A stone is projected at an initial speed of 120 ft/s directed 62 degrees above the horizontal , at a cliff of height h. The stone strikes the ground 5.5 s after launching. Find the height of the cliff, the speed of the stone just before impact at A and maximum height H reached above the ground

Pretty much, its ball thrown from ground and landing at a cliff.

Homework Equations

The Attempt at a Solution

I think I did it, but I am 100 % sure I got something wrong, here's what I did:

First I found V0y=v0 * sin 62 =32.3 m/s

Next I found the time, when the ball has reached its maximum height.
t=(v0y-Vy)/g=(32.3-0)/9.8=3.3 s
Found the max height
ymax=v0yt-(gt^2)/2=32.3 * 3.3 -(*0.5*9.8 * 3.3^2)= 53.229 m.

Now this is where I think I got it wrong.

Now we calculate the speed of the stone JUST BEFORE THE IMPACT at A.

We know that the time it took the stone to get from the maximum height to cliff is =5.3-3.3 = 2 s

V=-gt(because the velocity at the maximum height of the stone is 0)
V=-21.56 m/s
as we need speed, its absolute value aka =21.56 m/s

We found the speed of the stone, max height H, let's find how many meters did the stone travel from the maximum height to the cliff.

S(from max height to cliff)=gt^2 *0.5=4*0.5*9.8 = 19.6 m(cuz velocity at the maximum height is 0)

So that's it, answers are
S(from max height to cliff) =53.229-19.6=33.629 m
Speed is 21.56 m/s
S(max)=53.229 m

Where did I made a mistake? I know I did it somewhere because I missed something, help please!

You said the initial speed was 120 ft/s, but then use metres and g = 9.81 throughout.

Was the initial speed actually 120 m/s?

 

[wuffle]

I converted 120 ft/s to m/s,
120 ft/s > 36.576 m/s and I used that value in the problem.I realized now that I found the speed just as the stone reaches point A wrong.

Speed^2=Vy(21.56)^2+(V*Cos(62)^2)=21.56^2+36.576 * cos(62)=27.562^2
Speed=27.562 m/sEDIT:
I found two mistakes, 5.5 -3.3 obviously , not 5.3-3.3
S(from max height to cliff)=2.2^2*0.5*9.8=29

 

[PeterO]

I converted 120 ft/s to m/s,
120 ft/s > 36.576 m/s and I used that value in the problem.


I realized now that I found the speed just as the stone reaches point A wrong.

Speed^2=Vy(21.56)^2+(V*Cos(62)^2)=21.56^2+36.576 * cos(62)=27.562^2
Speed=27.562 m/s


EDIT:
I found two mistakes, 5.5 -3.3 obviously , not 5.3-3.3
S(from max height to cliff)=2.2^2*0.5*9.8=29

All good.

I probably would have used [as in your use, the minus sign with the g terms represents direction and allows just the value to be plugged in].

y=V_0yt-(gt²)/2, and used t = 5.5 to find out directly how high the cliff was.

V_y = V_0y - gt again with t = 5.5 to find the vertical velocity when it landed on the cliff,

V_y² = V_0y² - 2gy_max to find maximum height.

That way each calculation is independent of the other, and depends only on the original data provided, rather than something I had calculated [other than the two velocity components].

 

[wuffle]

Thank you very much!

I appreciate your help!