Question

A ball thrown horizontally from a height of 8.80 m hits the ground at a distance of 16.3 m. Calculate the initial speed of the ball.

Answer 1

vertical direction: h = \(h_0\) + \(v_yt +\frac{1}{2}gt^2\) \(v_y =0, h_0 =0\) therefore t = 1.22s (you calculate it, it's quite simple, right?)
to horizontal direction, the same, but \(v_x\) is what you need
\[l =l_0 +v t +\frac{1}{2}gt^2\] where \(l_0 =0 and g =0\)
so you have 16.3 = v t or v = 16.3/t = 16.3/1.22= 13.4m/s^2