1) The landing spot of the projectile is given by [tex]d=v_x \sqrt{\frac{2h}{g}}[/tex]
2) The common variable is the time
Explanation:
1)
The motion of a projectile consists of two independent motions:
- A uniform motion (constant velocity) along the horizontal direction
- A uniformly accelerated motion, with constant acceleration (acceleration of gravity) in the downward direction
The landing spot can be determined in the following way:
- First of all, we analyze the vertical motion to find the time of flight of the projectile. This can be done by using the suvat equation
[tex]h=ut+\frac{1}{2}at^2[/tex]
where
h is the vertical displacement of the projectile, which corresponds to the height from which the projectile has been fired, above the ground
u = 0 is the initial vertical velocity
[tex]a=g=9.8 m/s^2[/tex] is the acceleration of gravity
t is the time of flight
Solving for t,
[tex]t=\sqrt{\frac{2h}{g}}[/tex]
- After finding the time of flight, we analyze the horizontal motion, which is a uniform motion with constant horizontal velocity [tex]v_x[/tex]. Therefore, the horizontal distance covered is given by
[tex]d=v_x t[/tex]
And substituting the time of flight,
[tex]d=v_x \sqrt{\frac{2h}{g}}[/tex]
2)
Since the horizontal motion is uniform, the horizontal component of the displacement of the projectile is given by
[tex]x=v_x t[/tex]
where [tex]v_x[/tex] is the horizontal velocity and t is the time.
The vertical motion is accelerated, so the vertical component of the displacement is given by
[tex]y=\frac{1}{2}gt^2[/tex]
where g is the acceleration of gravity and t is the time.
Therefore, from the two equations we see that the common variable is t, the time.
Learn more about projectile:
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