MASSACHUSETTS INSTITUTE OF TECHNOLOGY
Department of Physics
8.01
W02D1_4 Table Problem: Initial and Final Conditions Projectile Motion
A physics instructor has set up a projectile gun a horizontal distance 5 m away from a
target that is 1 m above the ground. The instructor fires the projectile with an initial
speed 10 m  s 1 at an angle of 15o with respect to the horizontal. At the instant the
projectile leaves the gun, it is initially at the same height above the ground as the target.
Ignore air resistance. Let g  10 m  s 2 .
a) Write down a vector expression for the initial velocity of the projectile. Clearly
indicate your choice of coordinate system along with your choice of unit vectors.
b) Write down a vector expression for the position and velocity of the projectile
when it has returned to the same height that it was fired out.
Solution:
Understand the problem: A motion diagram and coordinate system is shown in the
figure below where we choose the origin at the point where the projectile leaves the gun.
The projectile has coordinates  x2 (t ), y2 (t )  . Because we are ignoring air resistance, the
projectile undergoes constant acceleration a y  g downwards, ax  0 .
Equations of motion for the projectile:
The initial position is given by x0  0 , y0  0 . The components of the initial velocity are
given by vy, 0  v0 sin(0 )  (10 m  s1 )sin(15o) and vx, 0  v0 cos(0 )  (10 m  s1 )cos(15o) ,
where v0  10 m  s1 is the magnitude of the initial velocity and 0  15o is the initial
angle with respect to the horizontal. The initial velocity is therefore
v 0  vx,0 î  vy,0 ĵ  (10 m  s1 )(cos(15o)î  sin(15o) ĵ)
The equations for position and velocity of the projectile simplify to
x(t)  v0 cos(0 )t
vx (t)  v0 cos(0 )
1 2
gt
2
v y (t)  v0 sin( 0 )  gt.
y(t)  v0 sin( 0 )t 
(1)
(2)
At the time t  t f , when the projectile returns to the initial height y(t f )  0 , we have that
0  v0 sin(0 )t f 
2v sin(0 )
1 2
.
gt f  t f  0
2
g
Therefore
2v0 sin( 0 )v0 cos( 0 ) v02 sin(2 0 )
x(t f )  v0 cos( 0 )t f 

.
g
g
Substituting our values for the given quantities yields
x(t f ) 
(10 m  s-1 )2 sin(30o)
 5 m.
10 m  s-2
So the projectile hits the target. The position vector at t  t f is r(t f )  (5 m)î . Because
the projectile has returns to the same height above the ground, the velocity has only
changed the sign of its y-component hence
v(t f )  (10 m  s1 )(cos(15o)î  sin(15o) ĵ) .