A soccer player kicks a ball from the origin with an initial speed of 45 m/s at

Question

A soccer player kicks a ball from the origin with an initial speed of 45 m/s at an angle of 40.0degree above the horizontal. The ball lands in a hole in the ground that is 10.0 m below the level where the ball was kicked. Neglect air resistance. Questions 13-17 refer to this situation. How long is the ball in the air? Less than 3 0 sec.; Between 3 0 and 6 0 sec.; Between 6 0 and 9 0 sec.; More than 9.0 sec. How far has the ball traveled in the horizontal direction when it lands? Less than 50 m; Between 50 and 100 m; Between 100 and 150 m; More than 150 m. What is the speed of the ball when it lands? Less than 45 m/s; Between 45 and 50 m/s; Between 50 and 55 m/s; More than 58 m/s. At the highest point of its path, the ball's speed is Zero; Equal to 45 cos40degree; Equal to 45 sin 40degree; Exactly 45 m/s At the highest point of its path, the net force acting on the ball is; Zero; Pointing in horizontal direction; Pointing straight down None of the above. Ignoring air resistance, a ball that is hit at an angle of 40degree will have the same range as a ball hit at an angle of: 20degree; 45degree; 50degree; None of the above If the initial speed of a projectile is doubled, all other quantities remaining the same, the range will: Halve; Remain the same; Double;

Explanation / Answer

time of flight = 2 * v * sin(theta) / g

time of flight = 2 * 45 * sin(40) / 9.8

time of flight = 5.9 sec

time it'll take to reach end point of the hole be t

10 = 45 * sin(40) * t + 0.5 * 9.8 * t^2

t = 0.32 sec

total time = 5.9 + 0.32

total time = 6.22 sec

so,

ball will be in the air for between 6 - 9 sec

horizontal range = v^2 * sin(2 * theta) / g

horizontal range = 45^2 * sin(80) / 9.8

horizontal rrange = 203.493

so,

more than 150 m

speed of the ball when it lands

v^2 = u^2 + 2as

v^2 = (45 * sin(40))^2 + 2 * 9.8 * 10

v = 32.135 m/s

less than 45 m/s

at heighest point speed = 45 * cos(40) m/s

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