(13%) Problem 5: A swimmer bounces almost straight up from a diving board and fa

Question

(13%) Problem 5: A swimmer bounces almost straight up from a diving board and falls vertically feet first into a pool. She starts with a speed of 4.2 m/s, and her takeoff point is 1.55 m above the pool's surface. 33% Part (a) For how long are her feet in the air, in seconds? 33% Part (b) What is the height above the board in meters, of her highest point? 33% Part (c) What is her velocity, in meters per second, when her feet hit the water? Assume the vertical component of the velocity is positive upwards

Explanation / Answer

Part (a) -

Use the expression -

x = xo + vot - a*t^2 / 2

put the values -
x = 1.55 + 4.2*t - (9.81/2)t^2 = 0

=> 1.55 + 4.2*t - 4.9*t^2 = 0

=> 4.9*t^2 - 4.2*t - 1.55 = 0

=> t = [4.2 + sqrt(4.2^2 + 4*4.9*1.55)] / (2*4.9) = [4.2 + 6.93] / 9.8 = 1.14 s

therefore, her feet are 1.14 s in the air.

Part (b) -  

Highest point: maximum -> calculate the derivative
x' = 4.2 - 9.81t = 0

=> t = 0.43
x = 1.55 + 4.2*0.43 - (9.81/2)* 0.43^2
x = 1.55 + 1.806 - 0.907 = 2.45 m

So, her highest point above the board is 2.45 - 1.55 = 0.90 m

Part (c) -

Use the expression -

v = vo - a*t
Now, from Part (a) we know the t = 1.14 s when she hits the water:
v = 4.2 - 9.81* 1.14 = - 6.98 m/s

therefore, she is falling with a velocity of 6.98 m/s.

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