Hi 2 Every1 Can Some1 Please Plaz Help Me?

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In summary: You know the final vertical speed is 25m/s down. The horizontal speed is Vcos(30), and the vertical speed is Vsin(30). You can use the vertical motion formula (y=Vt+1/2at^2) to find time (because vertical displacement is 22m, a=-9.8m/s^2, y=0, and V=-25m/s). Then you can use the horizontal motion formula (x=Vt) to find the horizontal displacement. Now that you have the displacement, time, and angle, you can use a trigonometric formula to find the initial velocity. In summary, the initial velocity of the stone thrown from the cliff can be found by breaking down
  • #1
SAIRA123
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HI TO ALL, THIS IS SAIRA, I AM NEW HERE.
I AM REALLY WORRIED AS I HAVE A PHYSICS ASSIGNMENT AND I AM REALLY STRUGGLING BECAUSE I MISSED ALOT OF LESSONS BECAUSE I WAS ILL.
COULD SOME1 PLZ HELP ME WITH THESE PROBLEMS, I WILL BE EVER SO GRATEFUL!

1. A boy throws a stone from the edge of a cliff with an initial speed V o at an angle of 30 degrees above the horizontal. When it hits the surface of a lake 22m below the top of the cliff, the stone is moving with a speed of 25 meter per second. What was its initial velocity?
 
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  • #2
SAIRA123 said:
HI TO ALL, THIS IS SAIRA, I AM NEW HERE.
I AM REALLY WORRIED AS I HAVE A PHYSICS ASSIGNMENT AND I AM REALLY STRUGGLING BECAUSE I MISSED ALOT OF LESSONS BECAUSE I WAS ILL.
COULD SOME1 PLZ HELP ME WITH THESE PROBLEMS, I WILL BE EVER SO GRATEFUL!

1. A boy throws a stone from the edge of a cliff with an initial speed V o at an angle of 30 degrees above the horizontal. When it hits the surface of a lake 22m below the top of the cliff, the stone is moving with a speed of 25 meter per second. What was its initial velocity?

Don't worry, it's actually very simple. Remember that horizontal and vertical motion are independant of each other (i.e. the vertical motion is affected by gravity, whereas horizontal motion is constant- that is, if there is no friction).
Split the initial and final speeds into their horizontal and vertical components.
 
  • #3


Hi Saira, welcome to the community! I'm sorry to hear that you've been ill and missed some lessons. Physics can definitely be a tricky subject, but don't worry, I'm sure we can help you with your assignment.

To answer your question, we can use the formula for projectile motion: vf^2 = vo^2 + 2ad. In this case, vf is 25 m/s, a is -9.8 m/s^2 (acceleration due to gravity), and d is 22 m. We can solve for vo by rearranging the formula:

vo^2 = vf^2 - 2ad
vo^2 = (25 m/s)^2 - 2(-9.8 m/s^2)(22 m)
vo^2 = 625 m^2/s^2 + 431.2 m^2/s^2
vo^2 = 1056.2 m^2/s^2
vo = √1056.2 m/s
vo ≈ 32.5 m/s

So the initial velocity of the stone was approximately 32.5 m/s at an angle of 30 degrees above the horizontal. I hope this helps! Let us know if you have any other questions or need further clarification. Good luck with your assignment!
 

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