A ball is launched from a height of 6 feet at a speed of 24 feet/second. If the ball follows projectile motion, how long will it take for the ball to touch the ground? (Hint: For projectile motion, height (h) = vt – 0.5gt2 + initial launch height, and acceleration due to gravity (g) = 32 feet/second2.)

Question

jdoe0001 · Answer

$\bf \textit{initial velocity}\
h = -16t^2+v_ot+h_o \qquad \text{in feet}\ \quad \
v_o=\textit{initial velocity, in this case }24\
h_o=\textit{initial height, in this case }6\ \quad \
h = -16t^2+v_ot+h_o\implies {\color{blue}{ h = -16t^2+24t+6}}$