Physics (Projectile motion) 2 (1 Viewer)

[atar90plus]

Hello

Sorry guys but I need help with 3 questions from the same handout from yesterday if you don't mind. I did not post it in the same thread because people may get confused on the questions I need help with.Please show full working and solution.

5. A ball is thrown from a bridge at v m/s , at an angle of 30 degrees from the horizontal, and travel out over the river below, landing in the water. The ball was recorded to take 2.5 seconds from when thrown, till it hit the water, travelling horizontal at a distance of 22.5m. Determine the speed of the ball. (Answer is 10.4m/s)

For this question I used the formula v=u+at but I got it wrong because my final answer is bigger than the correct answer

8. An asteroid miner of mass 80kg experiences a weigh force only 2% of that on earth

b) With this equipment, he could survive a fall with an impact velocity of 20m/s
i) Determine how long he could fall from the rest from a cliff on the asteroid and survive? (Ans: 102 sec)

ii) How high would the cliff be ? (Ans: 1.02km)

Please note the acceleration due to gravity from the asteriod is 0.196m/s

11. An object travelling with a horizontal velocity of 25m/s slides off a friction less bench 1.25m above the floor. How far from the bench does the object hit the ground? (Ans:12.63m)

For this question I found Ux = 25 and Uy= 0

 

[hayabusaboston]

Question: are u goin ahead from initiative or tutoring>?

 

[RivalryofTroll]

Question 5

We know that ux = ucos30 (30 degrees is the given angle)

We know that horizontal displacement (or delta x) = uxt

delta x = 22.5m (as given)
t = 2.5s given

We need to find u:

22.5 = ucos30 x 2.5
u = 22.5/(cos30 x 2.5)

Therefore u = 10.39 m/s

 

[RivalryofTroll]

Question 11

Vertical distance (or delta y) = 1/2gt^2

Delta y = 1.25 (given)
1.25 = 1/2(9.8)t^2
Therefore, t = 0.5050762723

We know that delta x = uxt

We know that ux = 25m/s (given)

Therefore delta x (or what we want to find which is the horizontal displacement) = 25 x 0.5050762723 = 12.63 m

 

[cineti970128]

i will post other questions later but for 8 b i

g accerelation constant = 0.196
Hence using v 2 =u 2 +2a Δy
Δy = (20)^2 / (2 X 0.196) = 1020.41m

Also using Δy = ut + 1/2 at^2 because u = 0
t^2 = (2 x Δy)/0.196

t=102 sec

 

[cineti970128]

Ha ^^ guess Rivarlyof Troll beated me in question 11
btw answer for 8 bii is part of my working ie 1020m

 

[RivalryofTroll]

i will post other questions later but for 8 b i

g accerelation constant = 0.196
Hence using v 2 =u 2 +2a Δy
Δy = (20)^2 / (2 X 0.196) = 1020.41m

Also using Δy = ut + 1/2 at^2 because u = 0
t^2 = (2 x Δy)/0.196

t=102 sec

This is impressive for a Year 10. Great job

 

[cineti970128]

Thanks

 

[atar90plus]

Thanks, Are you an accelerant for physics?

i will post other questions later but for 8 b i

g accerelation constant = 0.196
Hence using v 2 =u 2 +2a Δy
Δy = (20)^2 / (2 X 0.196) = 1020.41m

Also using Δy = ut + 1/2 at^2 because u = 0
t^2 = (2 x Δy)/0.196

t=102 sec

 

[atar90plus]

Thanks

Question 11

Vertical distance (or delta y) = 1/2gt^2

Delta y = 1.25 (given)
1.25 = 1/2(9.8)t^2
Therefore, t = 0.5050762723

We know that delta x = uxt

We know that ux = 25m/s (given)

Therefore delta x (or what we want to find which is the horizontal displacement) = 25 x 0.5050762723 = 12.63 m

 

[cineti970128]

no i just studied by myself

 

[someth1ng]

That's very good, you have a lot of potential (from what can be seen). You should be able to excel in physics!