Respuesta :
Answer:
a)α= 53.13°
b)The velocity at the highest point = Â 15 m/s
The acceleration at the highest point = 9.8[tex]m/s^2[/tex]
c)h=15 m
V=18.02 m/s
Explanation:
Speed of water ,u= 25 m/s
So the horizontal component of speed u = u cos α
Given that horizontal distance cover by water in 3 s is 45 m.
So We know that in projectile motion horizontal acceleration is zero.
In horizontal direction
Distance = Velocity x time
45 =  u cos α  x 3
u cos α = 45
45 = 25 cos α x 3
 cos α = 45/75
α= 53.13°
So the velocity at the highest point =  u cos α
The velocity at the highest point = Â 15 m/s
The acceleration at the highest point = 9.8[tex]m/s^2[/tex]
 Now the velocity along vertical direction(Vo) =  u sin α
   Vo= 25 sin 53.13°
Vo =20 m/s
[tex]h=V_o.t-\dfrac{1}{2}gt^2[/tex]
[tex]h=20\times 3-\dfrac{1}{2}\times 10\times 3^2[/tex]
h=15 m
So at 15 m above the ground water will strike .
The y-component of velocity after 3 sec
Vy= Vo - g t
Vy = 20 - 10 x 3
Vy= -10 m/s
The horizontal component of velocity will remain 15 m/s.
The resultant velocity
[tex]V=\sqrt{10^2+15^2}\ m/s[/tex]
V=18.02 m/s
The answers are as follows:
(a) The value of angle α is 53.13°
(b)The velocity at the highest point is 15 m/s and the acceleration at the highest point is 9.8m/s²
(c)The water strikes the building at a height of 15 m with a velocity of 18 m/s
Projectile Motion:
The given problem is a case of projectile motion. Now the data we have been provided is as follows.
Speed of water ,u= 25 m/s
The horizontal component of speed uₓ = ucosα
The horizontal distance is d= 45 m.
And time t = 3s
In horizontal direction
Distance = Velocity x time,
since there is no acceleration in the horizontal direction
[tex]d=u_x\times t\\\\45 = u cos\alpha \times3\\\\u cos \alpha = 45\\\\45 = 25 cos \alpha \times 3\\\\cos \alpha = 45/75\\\\\alpha= 53.13^o[/tex]
The velocity at the highest point will only be the horizontal velocity since at the highest point vertical velocity is zero, that is:
[tex]u_x=u\times cos53.13= 15m/s[/tex]
The acceleration at the highest point is g = 9.8m/s²
The vertical component of velocity
[tex]u_y=usin\53.13^o=25\times sin53.13\\\\u_y=20m/s[/tex]
Now using the second equation motion, the height h is given by:
[tex]h=u_yt-\frac{1}{2}gt^2\\\\\\h=20\times3-0.5\times9.8\times3^2\\\\h\approx15m[/tex]
So the water strikes at a height of 15m.
The horizontal velocity after 3 sec remains the same [tex]v_x=u_x=15m/s[/tex].
From the first equation of motion, the vertical velocity after 3 seconds becomes:
[tex]v_y= u_y - g t\\\\v_y = 20 - 10 \times 3\\\\v_y= -10 m/s[/tex]
So the resultant velocity after 3 sec will be;
[tex]v=\sqrt{v_x^2+v_y^2}\\\\v=\sqrt{15^2+(-10)^2}\\ \\ v=18m/s[/tex]
So the water strikes at a velocity of 18m/s.
Find out more about the equation of motion:
https://brainly.com/question/17007636
