Answer :
Missing question:
What is the initial speed of the shell?
Answer:
329.2 m/s
Explanation:
The motion of the shell is a projectile motion, consisting of:
- A uniform motion along the horizontal direction, with constant velocity [tex]v_x[/tex]
- A uniformly accelerated motion along the vertical direction, due to the force of graviity
We start by analyzing the vertical motion first. The shell falls down due to the effect of gravity, so we can use the following equation:
[tex]s=ut+\frac{1}{2}at^2[/tex]
where, taking downward as positive direction:
s = 80 m is the vertical displacement of the shell (= the height of the cliff)
u = 0 is the initial vertical velocity
t is the time of flight
[tex]a=g=9.8 m/s^2[/tex] is the acceleration of gravity
Solving for t, we find the time of flight:
[tex]t=\sqrt{\frac{2s}{a}}=\sqrt{\frac{2(80)}{9.8}}=4.04 s[/tex]
Now we know that along the horizontal direction, the shell travels a distance of
d = 1330 m
And it takes a time of
t = 4.04 s
We know that the horizontal velocity is constant, so we can find it as:
[tex]v_x = \frac{d}{t}=\frac{1330}{4.04}=329.2 m/s[/tex]
And since it is constant during the entire motion, this was also the initial horizontal velocity of the bullet.
Based on the data provided, the initial horizontal velocity is 329.2 m/s.
What is linear velocity?
Linear velocity is the horizontal displacement per unit time.
- velocity = displacement/time
The shell fired is a projectile.
Distance covered = 1330 m
To determine the initial horizontal velocity, the time taken for motion is calculated using the formula below:
[tex]t = \sqrt{ \frac{2h}{g} } [/tex]
where;
h = 80 m
g = 9.8 m/s
[tex]t \: = \sqrt{ \frac{2 \times 80}{9.8} } [/tex]
t = 4.04 s
Initial horizontal velocity = 1330/4.04
Initial horizontal velocity = 329.2 m/s
Therefore, the initial horizontal velocity is 329.2 m/s.
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