Hi = 1.6
u = 6.4 cos 65 forever = x coordinate
Vi = 6.4 sin 65 initial
Vertical problem:
v = Vi - 9.81 t
h = Hi + Vi t - 4.9 t^2
so
0 = 1.6 + 5.8 t - 4.9 t^2
or
4.9 t^2 - 5.8 t - 1.6 = 0
t = [ 5.8 +/- sqrt (33.64 + 31.36)]/9.8
t = (5.8 +/- 8.06)/9.8
we want + answer
t = 1.41 seconds in the aair
v = Vi - 9.81 t
v = 5.8 - 9.81 (1.41)
v = -8.08 m/s = y coordinate
A person who is 1.6 m tall throws a ball with a speed of 6.4 m/s at an angle of 65 degrees above the horizontal. What are the x and y components of the velocity (in m/s) of the ball the instant before it strikes the ground?
1 answer