(a) 1, (b) 3
Fig. 4-6, shows the direction of acceleration for a particle located
at its tail, and its length (representing the acceleration magnitude)
can be drawn to any scale.
Acceleration is second derivative of time, therefore 1
and 3 ax and ay both are constant. hence a is constant.
2 and 4 ay is constant but ax is not, therefore a is not
constant.
A special case of two-dimensional
motion: A projectile is an object
upon which the only force acting
is gravity.
Yes, see fig 4.9
Figure 4-9
The horizontal motion and the vertical motion are independent.
Figure 4-10 One ball is released
from rest at the same instant that
another ball is shot horizontally to
the right. Their vertical motions
are identical.
Eq. 2-15
Eq. 2-11
Eq. 2-16
horizontal distance
Solving (21) for x-xo=R, substituting y-yo=0 into (22) will give;
;
Solving (21) for t, substituting it into (22) will give;
Since
Therefore
V o=
Examples of Projectile Motion:
In the given picture below, Alice throws the ball to the +X direction with an
initial velocity 10m/s. Time elapsed during the motion is 5s, calculate the height
that object is thrown and Vy component of the velocity after it hits the ground.
-h=-1/2gt2
John kicks the ball and ball does projectile motion with an
angle of 53º to horizontal. Its initial velocity is 10 m/s, find the
maximum height it can reach, horizontal displacement and
total time required for this motion. (sin53º=0.8 and cos53º=0.6)
In the given picture you see the motion path of cannonball.
Find the maximum height it can reach, horizontal distance it
covers and total time from the given information. (The
angle between cannonball and horizontal is 53º and
sin53º=0, 8 and cos53º=0, 6)
Here xo and Ɵo = 0, to find t use;
y
x = -2m, v = - (4m/s)j
v
a
a = (4)2/2=16/2=(8m/s)i
y=2
v = - (4m/s)i
a = - (8m/s)j
a
v
x
The position of P with respect to A = Position of P with respect of B + position
of B with respect to A.
or sin70o
or cos70o
0
Ɵo=0
1.52 m
N
SE
E
SE
Vpg = Vpw + Vwg