- #1

CAT 2

- 44

- 1

## Homework Statement

A spring toy is launched from the ground at 2.3 m/s at an angle of 78° to the ground. What is the maximum height reached by the spring toy?

## Homework Equations

Δt = 2(V1)Sinθ/(acceleration)

ΔD = V1Δt + 1/2(acceleration)(Δt^2)

## The Attempt at a Solution

First I solved for time using this equation: Δt = 2(V1)Sinθ/(acceleration)

Δt = 2(2.3 m/s) Sin78 / (9.8 m/s)

Δt = 0.46s

Then I solved for distance using this equation: ΔD = V1Δt + 1/2(acceleration)(Δt^2)

ΔD = (2.3m/s)0.46s + 1/2(9.8 m/s)(0.46^2)

ΔD = 1.058 + 1.037

ΔD = 2.095m

Is this the right answer? Is this the right equation or should I be using ΔDv = 1/2(acceleration)Δt^2 ?Somehow this doesn't seem right because I don't know when the toy is at the highest point. If I knew the time of its highest point then I could find the vertical distance using that time, that is how I have been taught.

One more thing, should I be saying [up] is positive and using - 9.8 m/s[up] in my above equations?

e.g. ΔD = (2.3m/s[up])0.46s + 1/2(-9.8 m/s[up])(0.46^2)

Thanks for any help