Question

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Answer 1

1) The landing spot of the projectile is given by $d=v_x \sqrt{\frac{2h}{g}}$

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

$h=ut+\frac{1}{2}at^2$

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

$a=g=9.8 m/s^2$ is the acceleration of gravity

t is the time of flight

Solving for t,

$t=\sqrt{\frac{2h}{g}}$

- After finding the time of flight, we analyze the horizontal motion, which is a uniform motion with constant horizontal velocity $v_x$. Therefore, the horizontal distance covered is given by

$d=v_x t$

And substituting the time of flight,

$d=v_x \sqrt{\frac{2h}{g}}$

2)

Since the horizontal motion is uniform, the horizontal component of the displacement of the projectile is given by

$x=v_x t$

where $v_x$ is the horizontal velocity and t is the time.

The vertical motion is accelerated, so the vertical component of the displacement is given by

$y=\frac{1}{2}gt^2$

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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