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Sagot :
You are looking for Vx0 (initial velocity in X direction) use this formula
remember there is no acceleration in x direction, only Y
X = x0 + Vx0(t) + (1/2)at^2
X= 0 + Vx0 (t) + 0
x= Vx0 (t)
Vx0 = x/t
what we are missing is Time, so u can get this from using the Y formula to find time
Remember there is no velocity in Y direction
Y=y0 + Vy0(t) - (1/2)gt^2
y= 0 +0 - (1/2)gt^2
(3.0 m *2) / 9.80m/s/s = t^2
0.61224 = t^2
t = 0.78 s
now plug this time back into this formula from above
Vx0 = x/t
= 2.0m / 0.78s
=2.6
remember there is no acceleration in x direction, only Y
X = x0 + Vx0(t) + (1/2)at^2
X= 0 + Vx0 (t) + 0
x= Vx0 (t)
Vx0 = x/t
what we are missing is Time, so u can get this from using the Y formula to find time
Remember there is no velocity in Y direction
Y=y0 + Vy0(t) - (1/2)gt^2
y= 0 +0 - (1/2)gt^2
(3.0 m *2) / 9.80m/s/s = t^2
0.61224 = t^2
t = 0.78 s
now plug this time back into this formula from above
Vx0 = x/t
= 2.0m / 0.78s
=2.6
This question involves the concept of semi-projectile motion. It can be solved using the equations of motion in the horizontal and the vertical motion.
The minimum horizontal velocity required is "2.6 m/s".
First, we will analyze the vertical motion of the stunt person. We will use the second equation of motion in the vertical direction to find the time interval for the motion.
[tex]h=v_it+\frac{1}{2}gt^2[/tex]
where,
h = height = 3 m
vi = initial vertical speed = 0 m/s
t = time interval = ?
g = acceleration due to gravity = 9.81 m/s²
therefore,
[tex]3\ m = (0\ m/s)(t) + \frac{1}{2}(9.81\ m/s^2)t^2\\\\t^2 = \frac{(3\ m)(2)}{9.81\ m/s^2}\\\\t = \sqrt{0.611\ s^2}[/tex]
t = 0.78 s
Now, we will analyze the horizontal motion. We assume no air resistance, so the horizontal motion will be uniform. Hence, using the equation of uniform motion here:
[tex]s = vt\\\\v = \frac{s}{t}[/tex]
where,
s = horizontal distance = 2 m
t =0.78 s
v = minimum horizontal velocity = ?
Therefore,
[tex]v = \frac{2\ m}{0.78\ s}[/tex]
v = 2.6 m/s
Learn more about equations of motion here:
brainly.com/question/20594939?referrer=searchResults
The attached picture shows the equations of motion in the horizontal and vertical directions.

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