u = 16.1 cos 44 = horizontal velocity the whole time
Vi = 16.1 sin 44 = initial vertical velocity
h = Hi + Vi t - 4.9 t^2
3 = 0 + Vi t - 4.9 t^2
4.9 t^2 - (16.1 sin 44) t + 3 = 0
solve quadratic for t, use the big one
v = Vi - 9.8 t
u = 16.1 cos 44 still
speed^2 = u^2 + v^2
A golfer hits a shot to a green that is elevated 3 m above the point where the ball is struck. The ball leaves the club at a speed of 16.1 m/s at an angle of 44.0° above the horizontal. It rises to its maximum height and then falls down to the green. Ignoring air resistance, find the speed of the ball just before it lands.
1 answer
2 answers