A daring swimmer dives off a cliff with a running horizontal leap, as shown in t

Question

A daring swimmer dives off a cliff with a running horizontal leap, as shown in the figure. (Figure 1) What must her minimum speed be just as she leaves the top of the cliff so that she will miss the ledge at the bottom, which is 1.50 m wide and 9.50 m below the top of the cliff?Which of the following statements is the correct translation of the question statement into symbols? What is the diver's v0 when 0=0 and she is to land with x = 1.50 m and y = -9.50 m at time t? What is the diver's v 0 if she is to land with x0 = 1.50 m and y0 = -9.50 m at time t0? What is the diver's v0 when 0=0 and she is to land with x = 1.50 m and y = 9.50 m at time t? What is the diver's v 0 if she is to land with x = 1.50 m and y = -9.50 m at time t?

Explanation / Answer

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Answer: What is the diver's v0 when 0=0 and she is to land with x = 1.50 m and y = -9.50 m at time t?

Let us calculate Vo now

in horizontal direction:
x = 1.50 m
t = x/vo
= 1.50/vo ...eqn 1

in vertical direction:
a = -9.8 m/s^2
y = -9.50 m
vi = 0
use:
y = vi*t + 0.5*a*t^2
-9.50 = 0 + 0.5*(-9.8)*t^2
9.5 = 4.9*(1.50/vo)^2 {putting value of t from eqn 1}
vo^2 = 4.9*1.50^2 / 9.5
vo = 1.08 m/s
Answer: 1.08 m/s

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