- Joined
- Jul 22, 2011
- Messages
- 1,507
- Reaction score
- 318
A toy rocket is launched at an angle of 30° above the horizontal with an initial velocity of 10 m/s. If the rocket reaches a height of 10 m, how far away does it land from the platform?
(cos 30° = 0.866, sin 30° = 0.500)
Simple question.. How can it possibly reach a height of 10m when the initial velocity is 10m/s here?
Using vf^2 = vi^2 + 2aY, (vf = 0 at the top)
we know that Y = Vi^2/2g = (10sin30)^2/2g = 25/2g = 1.25m.. So the height should have been 1.25m instead of 10m....
I didnt even have to use the height for this question, but still, why the heck did they say it reached 10m when it is not even possible?
(cos 30° = 0.866, sin 30° = 0.500)
Simple question.. How can it possibly reach a height of 10m when the initial velocity is 10m/s here?
Using vf^2 = vi^2 + 2aY, (vf = 0 at the top)
we know that Y = Vi^2/2g = (10sin30)^2/2g = 25/2g = 1.25m.. So the height should have been 1.25m instead of 10m....
I didnt even have to use the height for this question, but still, why the heck did they say it reached 10m when it is not even possible?