A basketball is shot by a player at a height of 2.0m. The initial angle was 53° above the horizontal. At the highest point, the ball was travelling 6 m/s. If he scored (the ball went through the rim that is 3.00m above the ground), what was the player's horizontal distance from the basket?

At the ball's highest point, it has no vertical velocity, so the 6 m/s is purely horizontal. A projectile's horizontal velocity does not change, which means the ball was initially thrown with speed v such that

v cos(53°) = 6 m/s   ==>   v = (6 m/s) sec(53°) ≈ 9.97 m/s

The player shoots the ball from a height of 2.0 m, so that the ball's horizontal and vertical positions, respectively x and y, at time t are

x = (9.97 m/s) cos(53°) t = (6 m/s) t

y = 2.0 m + (9.97 m/s) sin(53°) t - 1/2 gt ²

Find the times t for which the ball reaches a height of 3.00 m:

3.00 m = 2.0 m + (9.97 m/s) sin(53°) t - 1/2 gt ²

==>   t ≈ 0.137 s   or   t ≈ 1.49 s

The second time is the one we care about, because it's the one for which the ball would be falling into the basket.

Now find the distance x traveled by the ball after this time:

x = (6 m/s) (1.49 s) ≈ 8.93 m

jupiter is the largest planet in our solar system, having a mass and radius that are respectively, 318 and 11.2 times that of earth. suppose that an object falls from rest near the surface of each planet and that the acceleration due to gravity remains constant during the fall. each object falls the same distance before striking the ground. determine the ratio of the time of fall on jupiter to that on earth.

where is the photo and the answer choices?