Physics 101 Group Session for Chapter 3
Group #______
1) Neglecting air resistance, is the acceleration of a projectile equal to zero when it reaches max height of its
trajectory? Explain your answer.
As long as air resistance is negligible, the acceleration of a projectile is constant and equal to the acceleration
due to gravity. The acceleration of the projectile, therefore, is the same at every point in its trajectory, and can
never be zero.
2) A bullet is fired from a rifle that is held 1.6 m above the ground in a horizontal position. The initial speed of
the bullet is 1100 m/s.
a) What is the time it takes for the bullet to strike the ground?
b) What is the range (horizontal distance) of the bullet?
REASONING AND SOLUTION
a. The bullet strikes the ground when its y coordinate is – 1.6 m. We know
voy = vosin 0° = 0
t
2 y
2  1.6 m 

 0.57 s
g
9.80 m/s2
b. During this time the bullet travels a horizontal distance x with ax = 0.
x = voxt = (vo cos o)t = (1100 m/s)(cos 0°)(0.57 s) = 630 m
3) On a spacecraft two engines fire for a time of 565 s. One gives the craft an acceleration in the x direction of
ax = 5.10 m/s2, while the other produces an acceleration in the y direction of ay = 7.30 m/s2. At the end of the
firing period, the craft has velocity components of vx = 3775 m/s and vy = 4816 m/s. What is the magnitude and
direction of the initial velocity? Express your answer as an angle with respect to the +x axis.
REASONING The magnitude and direction of the initial velocity v0 can be obtained using the
Pythagorean theorem and trigonometry, once the x and y components of the initial velocity v0x and v0y are
known. These components can be calculated using Equations 3.3a and 3.3b.
SOLUTION Using Equations 3.3a and 3.3b, we obtain the following results for the velocity components:
c
hb565 sg  893.5 m / s
 a t  4816 m / s  c 7 .30 m / s hb 565 s g  691.5 m / s
v 0 x  v x  a x t  3775 m / s  5.10 m / s 2
v0 y  v y
2
y
Using the Pythagorean theorem and trigonometry, we find
b893.5 m / sg  b 691.5 m / sg
F v I  tan FG 691.5 m / s IJ  37.7
GH v JK
H 893.5 m / s K
v 0  v 02 x  v 02 y 
  tan 1
0y
0x
2
1
2
 1130 m / s
4) To clear a 13.5 m barrier a cannon has to be launched at an angle of 15o with respect to the ground. What is
the launch speed of the cannon ball?
REASONING Since we know the launch angle  = 15.0, the launch speed v0 can be obtained using
trigonometry, which gives the y component of the launch velocity as v0y = v0 sin . Solving this equation
for v0 requires a value for v0y, which we can obtain from the vertical height of y = 13.5 m by using Equation
3.6b from the equations of kinematics.
SOLUTION From Equation 3.6b we have
v y2  v 02 y  2 a y y
b 0 m / sg
2
c
hb
 v 02 y  2 9 .80 m / s 2 13.5 m
g
or
c
Using trigonometry, we find
v0 
v0 y
sin 15.0

c
hb
2 9 .80 m / s 2 13.5 m
sin 15.0
hb
v 0 y  2 9 .80 m / s 2 13.5 m
g
62.8 m / s
g