# Another projectile motion problem.

-Dragoon-

## 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

Vv = vertical velocity, V1 = initial velocity, Dv = vertical distance.
Δt = 2V1Sinθ/(acceleration)
ΔDv = VvΔt + 1/2(acceleration)(Δt^2)
Vv = V1Sinθ

## The Attempt at a Solution

First I find the time, since I am given the initial velocity and the launch angle. I get a value of 0.5 seconds. Then I find the vertical velocity by using the equation above, and get about 2.24 m/s. Finally, I use all the information I calculated to find the vertical distance traveled by spring toy, and yield a negative value. What am I doing wrong? I am completely stumped as I haven't been taught any new equations to find the vertical distance. Thanks for all the help!

## Answers and Replies

dmouthfan2028
http://zonalandeducation.com/mstm/physics/mechanics/curvedMotion/projectileMotion/generalSolution/generalSolution.html" [Broken]

The vertical component of the initial velocity is $2.3sin(78^{\circ})$.
$V_{f} = v_{0} + at$
$t = \frac{V_{f} - v_{0}}{a} = \frac{0 - 2.3sin(78^{\circ})}{-9.81}$.

$D = v_{0}t + \frac{1}{2} at^{2} = (2.3sin(78^{\circ}))(\frac{0 - 2.3sin(78^{\circ})}{-9.81}) + \frac{1}{2}(-9.81)(\frac{0 - 2.3sin(78^{\circ})}{-9.81})^2 \approx .26$ meters.

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-Dragoon-
http://zonalandeducation.com/mstm/physics/mechanics/curvedMotion/projectileMotion/generalSolution/generalSolution.html" [Broken]

The vertical component of the initial velocity is $2.3sin(78^{\circ})$.
$V_{f} = v_{0} + at$
$t = \frac{V_{f} - v_{0}}{a} = \frac{0 - 2.3sin(78^{\circ})}{-9.81}$.

$D = v_{0}t + \frac{1}{2} at^{2} = (2.3sin(78^{\circ}))(\frac{0 - 2.3sin(78^{\circ})}{-9.81}) + \frac{1}{2}(-9.81)(\frac{0 - 2.3sin(78^{\circ})}{-9.81})^2 \approx .26 m$
I used the formula and still get the same negative value.

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