Answer:10.46s
Explanation:
To determine the time the cannonball is in the air, we can use the horizontal motion equation for projectile motion. The horizontal and vertical motions of a projectile are independent of each other.
The horizontal component of the initial velocity (
�
�
v
i
) can be found using trigonometry:
Horizontal component of
�
�
=
�
�
×
cos
(
angle
)
Horizontal component of v
i
=v
i
×cos(angle)
�
�
=
150
m/s
v
i
=150m/s (given)
angle
=
2
0
∘
angle=20
∘
So, the horizontal component of the initial velocity is:
�
�
=
150
m/s
×
cos
(
2
0
∘
)
v
x
=150m/s×cos(20
∘
)
�
�
≈
150
m/s
×
0.9397
v
x
≈150m/s×0.9397
�
�
≈
140.95
m/s
v
x
≈140.95m/s
The time the cannonball is in the air can be found using the vertical motion formula:
Vertical distance
=
�
�
×
sin
(
angle
)
×
�
−
1
2
�
�
2
Vertical distance=v
i
×sin(angle)×t−
2
1
gt
2
For the cannonball to land at the same height it was launched from, the vertical distance at the beginning and the end of the motion should be zero.
At the highest point of the cannonball's trajectory, its vertical velocity becomes zero. Using this fact, we can determine the time it takes to reach the peak of the trajectory and then double it to find the total time of flight.
The vertical component of the initial velocity (
�
�
�
v
iy
) is:
�
�
�
=
�
�
×
sin
(
angle
)
v
iy
=v
i
×sin(angle)
�
�
�
=
150
m/s
×
sin
(
2
0
∘
)
v
iy
=150m/s×sin(20
∘
)
�
�
�
≈
150
m/s
×
0.342
v
iy
≈150m/s×0.342
�
�
�
≈
51.3
m/s
v
iy
≈51.3m/s
The time to reach the peak (where
�
�
�
=
0
v
fy
=0) can be found using the formula:
�
�
�
=
�
�
�
−
�
×
�
v
fy
=v
iy
−g×t
Where
�
�
�
v
fy
is the final vertical velocity at the peak of the motion.
0
=
51.3
m/s
−
9.81
m/s
2
×
�
0=51.3m/s−9.81m/s
2
×t
�
=
51.3
m/s
9.81
m/s
2
t=
9.81m/s
2
51.3m/s
�
≈
5.23
s
t≈5.23s
Since the time to reach the peak is half the total time of flight, the total time in the air is
2
×
5.23
s
=
10.46
s
2×5.23s=10.46s
