Question

You kick a soccer ball with the speed of 31 M/S and angle of 50.how long does it take the ball to reach the top of its trajectory

Answer 1

do you know how to find the x-value of the vertex?

Answer 2

from a quadratic equation I mean

Answer 3

Judas, as simple as that is I doubt it will be sufficient, nor will they understand why. A teacher might mark them down for not showing enough work (going the long way).

We can start with our givens, initial velocity and the angle.
Now, because we know it is kicked at an angle it can be split up into two separate velocities, one in the x direction and one in the y.



Therefore, using what we know about trigonometry, we can figure out the x and y velocities.

\[\sin 50=\frac{ y velocity }{ 31 }\]

So, \[31 \sin 50=y velocity\]

\[y velocity=23.747 m/s\]



Now we need to find the x velocity, \[\cos 50=xvelocity/31\]

So, \[31 \cos 50=x velocity\]

\[x velocity=19.926 m/s\]

Now that we have our numbers, we need the change in time, which we can find using the displacement equation and factoring out ts.

So we know the displacement equation, \[\Delta x=Vo \Delta t+\frac{ 1 }{ 2 }a \Delta t^2\]
Where vo is initial velocity, a is acceleration, delta t is change in time.

\[\Delta y=(Vo)(\Delta t)+\frac{ 1 }{ 2 }g \Delta t^2\]

In order for this to work, we need to know that we're looking for when the ball is on the ground, delta y=0. Obviously at t=0 it's zero, but we need the second value at zero.

\[0=Vo \frac{ \Delta t }{ \Delta t } + \frac{ 1 }{ 2 }g(\frac{ \Delta t^2 }{ \Delta t })\]

Since there's a common delta t there we can take it out.
\[0=Vo+\frac{ 1 }{ 2 }g \Delta t\]

Simplify. \[0=31-4.9 \Delta t\]
\[31/4.9=6.327\]

Which is the entire path and the vertex is halfway between the start and endpoint. So divide 6.327 by 2=3.163 seconds.

Now this we can put intot he original delta y.

\[\Delta y=31(3.163)-4.9(3.163^2)=49 m up \in the air. \]

Since that's a lot of work, we could just do it his way and do the vertex formula, \[v=\frac{ -b }{ 2a }\] In the form \[y=ax^2+bx+c\]
So a is acceleeration due to gravity, b is velocity, and c is starting point.

-(31)/2(9.8)=3.163, and you'd put that into the displacement equation and calculate.