Physics (Projectile motion) (1 Viewer)

[atar90plus]

Hello

Could you guys please help me with these questions. I tried to do them but I did not get the correct answer. Please show full working and correct answer

1. A cannon ball fired at an unknown velocity (v) at an angle of 30 degrees from the horizontal. It reaches a maximum height of 490m.
a) Determine the initial velocity of the cannon ball?

For my working out I found that Ux=490cos30 = 424.35 and Uy = 490sin30 =245. Then I used u^2=v^2+2as to find the initial velocity
I just guessed this question so I believe I done it wrong

6. A cannon ball fired at an angle of 25 from the horizontal in the air for 16 seconds before falling back to the ground. What was its original speed?


The correct answers for Q1 a) is 196m/s and Q6 is 185.8m/s

 

[someth1ng]

Give me 15 minutes...

 

[atar90plus]

Ok Thanks

 

[someth1ng]

For question 1, I got 61.98ms-1 at 30 degrees against the horizontal. Is this correct?

v=u+at and s=ut+.5at^2 [use the notation in the exam]

v=u+at
0=u-9.8t
.'. u=9.8t

s(vertical)=u(vertical)t+.5at^2
490=9.8t^2+.5(-9.8)t^2
490=4.9t^2
t=10 seconds

Back to the v=u+at (vertical velocity)
0=u(vertical)-9.8*10
u(vertical)=98ms-1

v=98/sin30 ms-1
v=196mx-1(2sf)

 

[someth1ng]

For question 6, I got 185.51ms-1 at 25 degrees against the horizontal. Is this correct?

s=ut+0.5at^2
0=ut+0.5(-9.8)t^2

Since t does not equal to 0, t can be factorised out.

.'. 4.9t=u(certical)
u(vertical)=4.9x16
u(vertical)=78.4ms-1

v=78.4/sin25
v=185.51ms-1

 

[atar90plus]

I think so. For Q6 the answer is 185.8m/s and for Q1 the answer is 196m/s. The decimal point is a bit off so I am not sure. Could you please post the working out and solution please

 

[bleakarcher]

Hello

Could you guys please help me with these questions. I tried to do them but I did not get the correct answer. Please show full working and correct answer

1. A cannon ball fired at an unknown velocity (v) at an angle of 30 degrees from the horizontal. It reaches a maximum height of 490m.
a) Determine the initial velocity of the cannon ball?

For my working out I found that Ux=490cos30 = 424.35 and Uy = 490sin30 =245. Then I used u^2=v^2+2as to find the initial velocity
I just guessed this question so I believe I done it wrong

6. A cannon ball fired at an angle of 25 from the horizontal in the air for 16 seconds before falling back to the ground. What was its original speed?


The correct answers for Q1 a) is 196m/s and Q6 is 185.8m/s

I'll do a).

a) v(y)^2=u(y)^2+2a*s(y)=(v*sin(30))^2-2g*s(y)
When s(y)=490, v(y)=0, => v^2/4=2g*490
i.e. v=196 m/s taking g=9.8 m/s^2

 

[someth1ng]

I think so. For Q6 the answer is 185.8m/s and for Q1 the answer is 196m/s. The decimal point is a bit off so I am not sure. Could you please post the working out and solution please

I think they didn't use exact values for that one.

 

[someth1ng]

I just did a, 196 is correct lol, accidentally did root of 100 as root 10 LOL

 

[atar90plus]

I believe you done question 6 correctly. But for question one you did it wrong. Could you please post up the working out and solution for question 6 please

 

[atar90plus]

Is "g" positive or negative 9.8m/s^2 ?

I'll do a).

a) v(y)^2=u(y)^2+2a*s(y)=(v*sin(30))^2-2g*s(y)
When s(y)=490, v(y)=0, => v^2/4=2g*490
i.e. v=196 m/s taking g=9.8 m/s^2

 

[someth1ng]

I believe you done question 6 correctly. But for question one you did it wrong. Could you please post up the working out and solution for question 6 please

I just fixed both of my solutions.

 

[SpiralFlex]

For question 6, I got 185.51ms-1 at 25 degrees against the horizontal. Is this correct?

This is the right answer.

Are you using the dotpoint book? Atarplus

 

[bleakarcher]

Is "g" positive or negative 9.8m/s^2 ?

Sorry the solution is pretty poor but yeah g=-9.8 m/s^2 since I took the initial velocity as positive.

 

[atar90plus]

Ok. Thanks guys for your help. Now hopefully I can do the rest of the questions in my homework sheet

 

[atar90plus]

Ok

Sorry the solution is pretty poor but yeah g=-9.8 m/s^2 since I took the initial velocity as positive.

 

[dalonga]

PEAK hw?

 

[atar90plus]

Yes.

 

[atar90plus]

Got two handouts due this week

 

[atar90plus]

Hope that there is no quiz this week