Question

Question and answer in projectiles

Answer 1

Snowballs are thrown with a speed of 18 m/s from a roof 9.0 m above the ground. Snowball A is thrown straight downward; snowball B is thrown in a direction 30° above the horizontal. What is the direction of each snowball?

Projectile Motion:
An object thrown at an angle above the ground undergoes the motion experiencing the force resulting from gravity acting on it vertically. The object is called a projectile, and it follows a particular trajectory. When the angle is 90 degrees, then the trajectory is straight.

Answer and Explanation:


Given data:

The velocity with which the snowballs are thrown is,
u
=
18
m
/
s
.
The height of the roof above the ground is,
h
=
9.0
m
.
The snowball A is thrown straight downward.
The snowball B is thrown at an angle
θ
=
30
∘
.


The diagram can be visualised as:

Diagram of motion


For snowball A,

Consider the value of acceleration due to gravity as,
g
=
9.8
m
/
s
2
.

The angle of projection is zero. The velocity is in y-direction only as per the diagram and there is no velocity in x-direction.

So,

u
A
y
=
u
u
A
x
=
0


The snowball is under the effect of acceleration due to gravity. The velocity in y-direction just before landing is given by,

v
A
y
=
√
u
2
A
y
+
2
g
h
=
√
u
2
+
2
g
h


Substitute the known values,

v
A
y
=
√
(
18
m
/
s
)
2
+
2
×
9.8
m
/
s
2
×
9
m
=
22.37
m
/
s
v
A
y
=
(
18
m
/
s
)
2
+
2
×
9.8
m
/
s
2
×
9
m
=22.37
m
/
s


Since, there is no acceleration in x-direction. So, the velocity in x-direction just before landing is given by,

v
A
x
=
0
v
A
x
=
0

The direction of snowball A before landing is,

tan
α
=
v
A
y
v
A
x
=
22.37
m
/
s
0
α
=
tan
−
1
(
∞
)
=
90
∘
tan⁡α =
v
A
y
v
A
x
=
22.37
m
/
s
0
α =
tan
−
1
(
∞
)
=
90
∘


For snowball B,

The angle of projection is
θ
=
30
∘
θ
=
30
∘
. The horizontal and vertical component of velocity is

u
B
y
=
u
sin
θ
u
A
x
=
u
cos
θ
u
B
y
=usin⁡θ
u
A
x
=ucos⁡θ

The snowball is under the effect of acceleration due to gravity.

Let
H
H
be the height attained by snowball B above
h
h
. At height
H
H
, the velocity of the ball is zero, that is,
v
B
H
=
0
v
B
H
=
0
. Therefore, the height
H
H
is calculated as,

v
2
B
H
−
u
2
B
y
=
−
2
g
H
v
2
B
H
−
u
2
sin
2
θ
=
−
2
g
H
v
B
H
2
−
u
B
y
2
=−2gH
v
B
H
2
−
u
2
sin
2
θ =−2gH

Substitute the known values,

0
−
(
18
m
/
s
)
2
sin
2
30
∘
=
−
2
×
9.8
m
/
s
2
×
H
H
=
4.133
m
0−
(
18
m
/
s
)
2
sin
2
30
∘
=−2×9.8
m
/
s
2
×H H =4.133
m


During the downfall, the total distance to be covered is

d
=
h
+
H
=
9
m
+
4
.133
m
=
13
.133
m
d =h+H =9
m
+
4
.133
m
=13
.133
m


The initial velocity for the downward motion will be considered as zero. The final velocity is calculated as,

v
2
B
y
−
v
2
B
H
=
2
g
d
v
B
y
2
−
v
B
H
2
=
2
g
d

Substitute the known values,

v
2
B
y
−
0
=
2
×
9.8
m
/
s
2
×
13.133
m
v
B
y
=
√
257.4
m
2
/
s
2
=
16.04
m
/
s
v
B
y
2
−0 =2×9.8
m
/
s
2
×13.133
m
v
B
y
=
257.4
m
2
/
s
2
=16.04
m
/
s


Since, there is no acceleration in x-direction. So, the velocity in x-direction just before landing is given by,

v
B
x
=
u
B
x
=
18
m
/
s
cos
(
30
∘
)
=
15.59
m
/
s
v
B
x
=
u
B
x
=18
m
/
s
cos⁡
(
30
∘
)
=15.59
m
/
s


The direction of snowball B before landing is,

tan
β
=
v
A
y
v
A
x
=
16.04
m
/
s
15.59
m
/
s
β
=
tan
−
1
(
1.03
)
=
45.9
∘
tan⁡β =
v
A
y
v
A
x
=
16.04
m
/
s
15.59
m
/
s
β =
tan
−
1
(
1.03
)
=
45.9
∘




The direction of snowball A is vertically downward.

The direction of snowball B is
β
=
45.9
∘
β
=
45.9
∘
with the horizontal direction.