Respuesta :
Answer:
Distance, y = -2.287 meters
Explanation:
Given that,
The separation between tow buildings, d = 4 m
Velocity of the stuntman, v = 5 m/s
Angle with respect to the flat roof, [tex]\theta=15^{\circ}[/tex]
We know that the horizontal component of velocity is given by :
[tex]v_x=v\ cos\theta[/tex]
The velocity of an object is equal to the total displacement divided by total time taken as :
[tex]v=\dfrac{d}{t}[/tex]
t is the time of flight
[tex]t=\dfrac{d}{v_x\ cos\theta}[/tex]
[tex]t=\dfrac{4}{5\ cos(15)}[/tex]
t = 0.828 seconds
Let he will fall to d distance of y. Using the second equation to find it as :
[tex]y=vt+\dfrac{1}{2}at^2[/tex]
Here, a = -g
[tex]y=u_yt-\dfrac{1}{2}gt^2[/tex]
[tex]u_y[/tex] is the vertical component of velocity
[tex]y=u\ sin\theta t-\dfrac{1}{2}gt^2[/tex]
[tex]y=5\ sin(15)\times 0.828-\dfrac{1}{2}\times 9.8\times (0.828)^2[/tex]
y = -2.287 meters
So, he will fall at a distance of 2.287 meters. Hence, this is the required solution.
