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Problem 1

A diver springs upward with an initial speed of 1.8 m/s from a 2.5 m board.

(a) Find the velocity with which he strikes the water. (Hint: When the diver reaches the water, his displacement is y = -2.5 m (measured from the board), assuming that the downward direction is chosen as the negative direction.)

(b) What is the highest point he reaches above the water?

The four kinematics equations

(a) used V^2 = V(initial)^2 + 2ay

V^2 = 1.8^2 + 2(-9.8)(-2.5) = 7.2 m/s which is wrong.

(b) Found out (b) to be 2.7 by 0 = 3.24 +19.6y and adding that to 2.5

Just can't seem to figure out (a). There is another problem which is similar and I guess I just don't seem to understand when something is shot/thrown up first and then falling.

Problem 2

A ball is thrown upward from the top of a 55.0 m tall building. The ball's initial speed is 12.0 m/s. At the same instant, a person is running on the ground at a distance of 37.0 m from the building. What must be the average speed of the person if he is to catch the ball at the bottom of the building?

The four kinematics equations

The time the ball is thrown up

V = V(initial) + (a)(t)

0 = 12.0 + 9.8(t)

t = 1.2245 s

The time the ball is thrown down

y = V(initial)*t + 1/2(a)(t^2)

t = 3.35

Total time of 1.2245 s + 3.35 s = 4.6 s

Avg. Spd = D/T

37 m / 4.6 = 8.0 m/s = not the right answer

## Homework Statement

A diver springs upward with an initial speed of 1.8 m/s from a 2.5 m board.

(a) Find the velocity with which he strikes the water. (Hint: When the diver reaches the water, his displacement is y = -2.5 m (measured from the board), assuming that the downward direction is chosen as the negative direction.)

(b) What is the highest point he reaches above the water?

## Homework Equations

The four kinematics equations

## The Attempt at a Solution

(a) used V^2 = V(initial)^2 + 2ay

V^2 = 1.8^2 + 2(-9.8)(-2.5) = 7.2 m/s which is wrong.

(b) Found out (b) to be 2.7 by 0 = 3.24 +19.6y and adding that to 2.5

Just can't seem to figure out (a). There is another problem which is similar and I guess I just don't seem to understand when something is shot/thrown up first and then falling.

Problem 2

## Homework Statement

A ball is thrown upward from the top of a 55.0 m tall building. The ball's initial speed is 12.0 m/s. At the same instant, a person is running on the ground at a distance of 37.0 m from the building. What must be the average speed of the person if he is to catch the ball at the bottom of the building?

## Homework Equations

The four kinematics equations

## The Attempt at a Solution

The time the ball is thrown up

V = V(initial) + (a)(t)

0 = 12.0 + 9.8(t)

t = 1.2245 s

The time the ball is thrown down

y = V(initial)*t + 1/2(a)(t^2)

t = 3.35

Total time of 1.2245 s + 3.35 s = 4.6 s

Avg. Spd = D/T

37 m / 4.6 = 8.0 m/s = not the right answer

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