# Another projectile motion problem.

• -Dragoon-
What am I doing wrong?In summary, the spring toy is launched from the ground at 2.3 m/s at an angle of 78° to the ground. By using the equations for vertical velocity and distance, the maximum height reached by the spring toy is approximately 0.26 meters. However, there may be an error in calculation as the resulting value is negative. Further review and clarification of the equations may be necessary.

## 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!

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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dmouthfan2028 said:
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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