Charlie Brown kicks a ball at 22.5 m/s at 33.0 degrees. How far has the ball moved horizontally when it reaches its highest point?
+33653 Initial vertical velocity uh = 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 uh = 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.
+14986 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 vy is zero at the highest point.
\(v_0\times sin\ \alpha-g\times t =0\)
\(\LARGE t= \frac{v_0\times sin \ \alpha}{g}\) \(\LARGE= \frac{22.5\frac{m}{s}\times sin \ 33°}{9.81\frac{m}{s^2}}\)
\({\color{blue}\large t = 1.249 \ s}\)
\({\color{blue}The \ ball \ has \ reached \ the \ highest \ point}\)
\({\color{blue}of \ the \ throwing \ parcel \ after \ 1.249 s.}\)
\(w= v_h \times t =18.87\ \frac{m}{s}\times 1.249\ s\)
\({\color{blue}\large w=23.569\ m}\)
\({\color{blue}The \ ball \ has \ a \ horizontal \ distance \ of \ 23.569\ m }\)
\({\color{blue}when \ it \ is \ at \ the \ highest \ point }\)
\({\color{blue}of \ the \ throwing \ parabola.}\) 
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