Projectile Motion

Initial vertical velocity u_h = 22.5*sin(33deg) m/s

At the highest point the vertical velocity is zero, so using v = u + at we have:

0 = 22.5*sin(33deg) - 9.8t. where the acceleration of gravity is 9.81m/s² downwards.  From this you can get the time, t, in seconds.

The horizontal velocity is given by u_h = 22.5*cos(33deg) m/s so the distance (in metres) travelled in time t is given by multiplying this velocity by the value found for t.

Charlie Brown kicks a ball at v₀ = 22.5 m/s at α = 33.0 degrees. How far has the ball moved horizontally when it reaches its highest point?

I decompose v₀ into the vertical and horizontal component.

$v_{h} = v_0\times cos \alpha=22.5\frac{m}{s}\times cos\ 33°= 18.87\frac{m}{s}$

$v_{v}= v_0\times sin\ \alpha = 22.5\frac{m}{s}\times sin\ 33°=12.2544\ \frac{m}{s}$

The vertical component v_y is zero at the highest point.

$v_0\times sin\ \alpha-g\times t =0$

$\LARGE t= \frac{v_0\times sin \ \alpha}{g}$ $\LARGE= \frac{22.5\frac{m}{s}\times sin \ 33°}{9.81\frac{m}{s^2}}$

${\color{blue}\large t = 1.249 \ s}$

${\color{blue}The \ ball \ has \ reached \ the \ highest \ point}$

${\color{blue}of \ the \ throwing \ parcel \ after \ 1.249 s.}$

 $w= v_h \times t =18.87\ \frac{m}{s}\times 1.249\ s$

${\color{blue}\large w=23.569\ m}$

${\color{blue}The \ ball \ has \ a \ horizontal \ distance \ of \ 23.569\ m }$

${\color{blue}when \ it \ is \ at \ the \ highest \ point }$

${\color{blue}of \ the \ throwing \ parabola.}$                   !