Answer:
Explanation:
Suppose v is the initial velocity and [tex]\theta[/tex] is the angle of inclination
distance traveled in vertical direction in t=1 s
When gravity is present
[tex]y=vt+\frac{1}{2}at^2[/tex]
where [tex]y=vertical\ distance[/tex]
[tex]a=acceleration[/tex]
[tex]t=time[/tex]
[tex]v=initial\ velocity[/tex]
here initial velocity is v\sin \theta [/tex] so
[tex]y=v\sin \theta \times 1-\frac{1}{2}gt^2[/tex]
[tex]y=v\sin \theta -0.5g[/tex]
In absence of gravity
[tex]y_2=v\sin \theta \times t[/tex]
[tex]y_2=v\sin \theta \times 1[/tex]
[tex]\Delta y=y_2-y=v\sin \theta -v\sin \theta +0.5 g=4.9\ m[/tex]
The projectile will fall below the line by 5 meters if it has been traveling for 1 second.
Recall that t=1s,
Thus, 1/2g (t ^2)
= 0.5 (9.8) (1 ^ 2)
= 4.9m
Which is approximately 5 m.
A projectile is any object that once released from a point above the ground or resting place continues in motion by its own inertia and is acted upon only by the downward force of gravity.
Learn more about Projectiles at:
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