Asked by Erik
A ball starts 1.175 meters off the ground and shot at an angle of 40 degrees with a velocity of 3.2 m/s. Where does it land?
Answers
Answered by
Damon
u = horizontal velocity = 3.2 cos 40 forever
Vi = initial up velocity component = 3.2 sin 40
v = Vi - 9.8 t
h = 1.175 + Vi t - 4.9 t^2
at ground, h = 0
so
4.9 t^2 - Vi t - 1.175 = 0
solve quadratic for t, use positive solution
then
range = u t
Vi = initial up velocity component = 3.2 sin 40
v = Vi - 9.8 t
h = 1.175 + Vi t - 4.9 t^2
at ground, h = 0
so
4.9 t^2 - Vi t - 1.175 = 0
solve quadratic for t, use positive solution
then
range = u t