A swimmer bounces straight up from a diving board and falls feet first into a pool. She starts with a velocity of 3.00 m/s, and her takeoff point is 1.10 m above the pool.
(a) How long are her feet in the air?
s
(b) What is her highest point above the board?
m
(c) What is her velocity when her feet hit the water?
m/s
To answer these questions, we can use the kinematic equations of motion. We'll assume that the acceleration due to gravity is approximately 9.8 m/s², and consider upwards as the positive direction.
(a) To find how long her feet are in the air, we can use the equation:
Δy = v_iy * t + (1/2) * a_y * t²,
where Δy is the vertical displacement or the height above the pool, v_iy is the initial vertical velocity (3.00 m/s, upwards in this case), a_y is the vertical acceleration (-9.8 m/s², downwards due to gravity), and t is the time in seconds.
Since she starts from a height of 1.10 m above the pool, her vertical displacement (Δy) is -1.10 m (taking downwards as negative).
Plugging in the values into the equation, we have:
-1.10 = 3.00 * t - (1/2) * 9.8 * t².
Rearranging the equation to standard quadratic form, we get:
-4.9 * t² + 3.00 * t - 1.10 = 0.
To solve for 't,' we can use the quadratic formula:
t = (-b ± √(b² - 4ac)) / 2a,
where a = -4.9, b = 3.00, and c = -1.10.
Plugging in the values, we find two possible values for 't': t ≈ 0.584 s and t ≈ 0.415 s.
However, since our problem only considers the time when her feet are in the air, we discard the smaller value of 't.'
Therefore, her feet are in the air for approximately 0.584 seconds.
(b) To find her highest point, we can use the equation for the vertical displacement:
Δy = v_iy * t + (1/2) * a_y * t².
At the highest point, her vertical velocity is zero (v_iy = 0). So the equation becomes:
Δy = (1/2) * a_y * t².
Plugging in the values, we have:
Δy = (1/2) * (-9.8) * (0.584)².
Evaluating the expression, we find:
Δy ≈ -1.9808 m.
Since upwards displacement is considered positive, the highest point above the board is approximately 1.9808 meters.
(c) To find her velocity when her feet hit the water, we can use the equation for final velocity:
v_fy = v_iy + a_y * t.
Plugging in the values, we have:
v_fy = 3.00 + (-9.8) * 0.584.
Evaluating the expression, we find:
v_fy ≈ -2.378 m/s.
Therefore, her velocity when her feet hit the water is approximately 2.378 m/s, downwards.