Rank these throws based on the maximum height reached by the ball.?
Rank these throws based on the maximum height reached by the ball.
Rank from largest to smallest. To rank items as equivalent, overlap them:
15M/S at 60 deg
10M/S at 90deg
15M/S at 45 deg
15 M/S at 30 deg
10 M/S at 60 deg
20M/S at zero deg
I ranked them
15M/S at 60 deg 1
10M/S at 90deg 3
15M/S at 45 deg 2
15 M/S at 30 deg 5
10 M/S at 60 deg 4
20M/S at zero deg4
and that was wrong any help would be much appreciated
There are only 5 slots, so two will have to overlap
The maximum height of the ball will depend on the vertical component of the velocity according to the equation
Vfy² = Viy² − 2*g*y
where
Viy is the original upward component of the velocity of the ball,
Vfy is the upward component of the velocity of the ball at maximum height, which will be zero,
g is the acceleration of gravity, which is 9.80 to 9.81 m/sec², depending on where the surface of the earth the experiment is performed,
and y is the maximum height the ball reaches above its initial position.
If you need the actual height of the ball, you can solve for y algebraically to give
y = Viy² / (2*g)
but you don’t really need to calculate the necessary heights. All you need to recognize is that the height will be proportional to the square of the upward component of the initial velocity Viy², which is the square of the initial speed V² times the square of the sine of the angle at which the ball is being thrown. In this problem, you are dealing only with the special angles zero, 30 deg, 45 deg, 60 deg, and 90 deg, which have sines of In this problem, you are dealing only with the special angles zero, 30 deg, 45 deg, 60 deg, and 90 deg, which have the squares of sines of
sin² (0 deg) = 0
sin² (30 deg) = 1/2² = 1/4
sin² (45 deg) = (√2/2)² = 1/2
sin² (60 deg) = (√3/2)² = 3/4
sin² (90 deg) = 1
Viy² is given by the square of initial speed of the ball times the square of the sine of the angle from horizontal. For example, the square of the initial velocity of the ball thrown at 15 m/s at 60 deg would be
[15 m/sec * sin (60 deg)]²
= (15 m/sec)² * sin² (60 deg)
= 225 m²/sec² * 3/4
= 168¾ m²/sec²
You have ranked all of the answers correctly except the ball thrown at 20 m/sec at 0 deg, i.e., horizontally. This throw has no positive vertical component and won’t rise above its initial height at all and will come in at #6. There are no ties in the rankings despite what the answer sheet may seem to imply.
we are approaching to zero degrees or to zero of iran negotiations-
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