Question

A soccer player kicks a ball with a speed of 30 m/s at an angle of 10. How long does the ball stay in the air?

Answer 1

since were not interested in how far the ball went, we can focus in on the y (vertical) axis.

We only want the vertical component of the velocity:


use some trig: \(sin(10^o)=\dfrac{v_y}{30~m/s} \rightarrow v_y=sin(10^o)*30~m/s=5.21~m/s\)

I'm gonna drop the y subscript.

Because it's a projectile, i.e. the trajectory is a parabola and thus \(v=0\) at the highest point and \(v=-v_i\) at the same point you started (the ground).

we can use the equation: \(v_f=v_i+a\Delta t\)

where \(v_f=-v_i\) because the velocity is in the opposite direction.

\(-v_i=v_i+a\Delta t\rightarrow\Delta t=\dfrac{-v_i-v_i}{a}\)

acceleration is due to gravity in the -y direction and \(v_i=5.21~m/s\)

\(\Delta t=\dfrac{-5.21~m/s-5.21~m/s}{-9.8~m/s^2}=\dfrac{2(5.21~m/s)}{9.8~m/s^2}=1.1~s\)

This seems like a really short time, but it was kicked at an angle of \(10^o\) which is really small, so it seems right.