I. Motion of a projectile
The velocity vector changes in both magnitude and direction.
A. Let v0 be the initial velocity and q 0 the angle between the horizontal and the vector v0
v0 = v0xi + v0yj
v0x = v0cosq 0 and v0y = v0sinq 0
B. Use vector equations: v = v0 + at and r - r0 = v0t + (1/2)at2
vx = v0x + axt and x - x0 = v0xt + (1/2)ax t2
vy = v0y + ayt and y - y0 = v0yt + (1/2)ay t2
C. For projectile motion:
1.ax = 0
2. ay = -g = -9.8 m/s2 = -32 f/ s2
3. vx = v0cosq 0
4. vy = v0sinq 0 -gt
5. Then: v2 = vx2 + vy2 and tanq = vy/vx
where q is the angle that the velocity vector makes with the horizontal at any time and v is the magnitude of the velocity vector at any time.
D. The projectile's coordinates, assuming that it starts from the origin are:
1. x = v0xt = (v0cosq 0)t
2. y = (v0sinq 0)t - (1/2)gt2
A soccer player kicks a ball at an angle of 37 degrees from the horizontal with an initial speed of 20 m/s.
a) Find the time to reach the highest point of the trajectory, T.
b) How high does the ball go?
c) What is the horizontal range of the ball?
d) How long is it in the air?
e) What is the VELOCITY of the ball when it strikes the ground?