moreemma
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In a movie production, a stunt person must leap from a balcony of one building to a balcony 3.0 m lower on another building. If the buildings are 2.0 m apart, what is the minimum horizontal velocity the stunt person must have to accomplish the jump? Assume no air resistance and that ay = −g = −9.81 m/s2 . (Ans. 2.6m/s) PLS SHOW WORK

Sagot :

danielagm210
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
hamzaahmeds

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.

View image hamzaahmeds
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