A basketball player throws a ball at 60° to the horizontal at a net 4.0m away. The height of the net is 3.3m. The ball leaves hi
s hands at a height 2.0m from the ground. What is the initial speed of the ball?
Th pic above is my ans, but it's quite wierd. So if it's wrong, what is the correct ans, and why my ans is wrong(what mistakes i did). Thanks
Th pic above is my ans, but it's quite wierd. So if it's wrong, what is the correct ans, and why my ans is wrong(what mistakes i did). Thanks
1 answer:
Excellent work!
Your calculations are correct, but near the end, you have forgotten to cancel u. If you cancel the u, the linear term becomes a constant, and the resultant equation becomes a simple quadratic which is much easier to solve.
I get u=7.468 m/s using g=9.81 (as you did)
You will rework and should get 7.468 m/s as well.
Congrats for the good work!
Your calculations are correct, but near the end, you have forgotten to cancel u. If you cancel the u, the linear term becomes a constant, and the resultant equation becomes a simple quadratic which is much easier to solve.
I get u=7.468 m/s using g=9.81 (as you did)
You will rework and should get 7.468 m/s as well.
Congrats for the good work!
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Answer:
1.8 × 10⁻⁸ Hm
Explanation:
Given that:
The refractive index of the film = 19
The wavelength of the light = 136.8 μ m
The thickness can be calculated by using the formula shown below as:
Where, n is the refractive index of the film
is the wavelength
So, thickness is:
Thickness = 1.8 μ m
Since,
1 μ m = 10⁻⁸ Hm
So,
Thickness = 1.8 × 10⁻⁸ Hm