The discussion shows that the motion of a ball that is projected at some angle can be analysed by separately looking at the horizontal and the vertical directions. All we have to do is consider two motions, one in the horizontal direction in which there is no acceleration, and another in the vertical direction in which we have an acceleration, g. Consider Figure 2.16, where a projectile is launched at an angle to the horizontal with speed u. The components of the initial velocity vector are ux = u cos and uy = u sin . At some later time t the components of velocity are vx and vy. In the x-direction we do not have any acceleration and so:
u uy
θ
ux
Figure 2.16 A projectile is launched at an angle to the horizontal with speed u.
vx = ux vx = u cos In the y-direction the acceleration is −g and so: vy = uy − gt vy = u sin − gt The green vector in Figure 2.17a shows the position of the projectile t seconds after launch. The red arrows in Figure 2.17b show the velocity vectors.
y /m 20
y /m 20 P
15 10 5
15 10
y
5 x
0 –5
0.5
1.0
1.5
0
x /m 2.0
a
–5
0.5
1.0
1.5
x /m 2.0
b
Figure 2.17 a The position of the particle is determined if we know the x- and y-components of the position vector. b The velocity vectors for projectile motion are tangents to the parabolic path.
Exam tip All that we are doing is using the formulas from the previous section for velocity and position v = u + at and s = ut + 12at2 and rewriting them separately for each direction x and y. In the x-direction there is zero acceleration and in the y-direction there is an acceleration −g.
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We would like to know the x- and y-components of the position vector. We now use the formula for position. In the x-direction: x = uxt x = ut cos And in the y-direction: y = uyt − 12gt2 y = ut sin − 12gt2