Problem 10: A swimmer bounces almost straight up from a diving board and falls v

Question

Problem 10: A swimmer bounces almost straight up from a diving board and falls vertically feet first into a pool. She starts with a speed of 3.4 m/s, and her takeoff point is 1.6 m above the pool's surface. Part (a) For how long are her feet in the air, in seconds? Numeric : A numeric value is expected and not an expression. Part (b) What is the height above the board, in meters, of her highest point? Numeric : A numeric value is expected and not an expression. 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 Numeric : A numeric value is expected and not an expression. vr

Explanation / Answer

Given,

v0 = 3.4 m/s ; d = 1.6 m

a)We know that

Y = Y0 + v0 t + 1/2 at^2

4.9 t^2 - 3.4 t - 1.6 = 0

t = 1.02 s

Hence, t = 1.02 s

b)We know that, at max height v = 0

v^2 = u^2 + 2 a S

0 = 3.4^2 + 2 x -9.8 x H => H = 0.59 m

Hence, H = 0.59 m

c)v = u + at

v = 3.4 - 9.8 x 1.02 = -6.6 m/s

Hence, v = -6.6 m/s (its negative as its directed downwards)

Related Questions

Navigate

• Browse (All)
• Browse P
• Subjects

---