# Problem 4-11 - Free fall - Part 11 - (C)

A ball is thrown from a balloon with an initial unknown velocity. The ball accelerates at $\mathrm{9.80\;m/s^2}$ downward for $\mathrm{2.00\; s}$, at which time its instantaneous velocity is $\mathrm{24.0\; m/s}$ at an angle of $\mathrm{45.0^\circ}$ below the horizontal. Determine the magnitude and direction of the initial velocity.

Accumulated Solution

$\mathrm{v = v_0 + at}$

$\mathrm{v_x = v_{0x} + a_xt \\ v_y = v_{0y} + a_yt \\ a_x = 0 \\ a_y = g \\ v_x = v_{0x} \\ v_y = v_{0y} + 9.8t \\ v_x = 24 \cos45 = 17.0\; m/s \\ v_y = 24 \cos45 = 17.0\; m/s \\ v_{0x} = v_{0x} = 17.0 \; m/s} \\ v_{0y} = -2.6 \; m/s \\ v_0 = 17.2 \; m/s$

No. No $\mathrm{v_x}$ is greater than $\mathrm{v_y}$ not less.

Try again.