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.

Question

loser66 · Answer

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 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