Based on the information given:
The vertical velocity is 13.9 * sin (35°) = 7.97 m/s
The horizontal velocity is 13.9 * cos (35°) = 11.39 m/s
First, let's see how long the ball will be in the air, given its initial vertical velocity of 7.97 m/s. To do this, let's see when the vertical velocity equals zero (when the ball reaches the vertical peak of its trajectory, or the moment just before the ball starts to come down). Remember that acceleration due to gravity is -9.8 /s²
v(t) = vi - at
0 = 7.97 + -9.8 * t Subtract 7.97 from both sides
-7.97 = -9.8 * t Divide both sides by -9.8
t = .8135
The height at this point was:
h(t) = vi (t) + (1/2) a *t²
h(t) = 7.97 (.8135) + (1/2) (-9.8) (.8135²)
h(t) = 6.48 - 3.24
h(t) = 3.24m
How long until the ball came down? How long does it take a ball to fall 3.24 meters, with an initial vertical velocity of 0?
3.24 = 0 - (1/2) (9.8) (t²) Divide both sides by (9.8/2)
t² = .66 Take the square root of both sides
t = .835
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The ball took about the same amount of time, .8 seconds, to come down as it did to come up. Total time ball was in air: .813 + .835 = 1.648
The balled traveled for 1.648 seconds at a constant horizontal speed (11.39 m/s) before coming down. How far did it travel?
1.648 * 11.39 = 18.77 m
What is the change given this new angle, compared to the 25° angle?
18.77 - 15.1 = 3.67
How do the two angles differ in terms of the general path of the ball? In the 25° angle, the ball was kicked at a shallower angle (kind of like a line drive hit in baseball), and traveled with a higher horizontal velocity and lower initial vertical velocity. This means the ball was in the air for less time. the 35° angle had a higher initial vertical and lower horizontal velocity, but was in the air for longer. This extra hang time means that the ball was able to travel farther before hitting the ground.
