Answer:
a) [tex]x=26m[/tex]
b) [tex]v_{y}=-21.5m/s[/tex]
Explanation:
From the exercise we know initial velocity, initial height
[tex]y_{o}=20m[/tex]
[tex]v_{o} =12m/s[/tex] [tex]\beta =45[/tex]º
a) The range of the stone is defined by how far does it goes. From the theory of free falling objects, we have:
[tex]y=y_{o}+v_{oy}t+\frac{1}{2}gt^{2}[/tex]
The stone strike the water at y=0
[tex]0=20+12sin(45)t-\frac{1}{2}(9.8)t^{2}[/tex]
Solving for t, using the quadratic formula
[tex]t=-1.33s[/tex] or [tex]t=3.06s[/tex]
Since time can't be negative, the answer is t=3.06s
Now, we can calculate the range of the stone
[tex]x=v_{o}t=(12cos(45))m/s(3.06s)=26m[/tex]
b) We can calculate the velocity were the stone strike the water using the following formula
[tex]v_{y}=v_{oy}+gt=12sin(45)m/s-(9.8m/s^{2})(3.06s)=-21.5m/s[/tex]
The negative sign indicates that the stone is going down
Answer:
(a) x= 26m
(b) vf= 23.15m/s
Explanation:
Given data
h=20m
Ɵ=45°
to find
(a) range of stone=x=?
(b) velocity=vf=?
Solution
For part (a)
You need to solve for time first using
yf = yi + visinƟt + 1/2gt^2
0 = 20m + 12sin45t + 1/2(-9.8)t^2
and use the quadratic equation to solve for t
t = 3.064 sec
To solve for the distance traveled use
x = xi + vicosƟt + 1/2at^2 there is no acceleration in the x direction so that cancels
x = 12cos(45)(3.064)
x= 26m
For part(b)
For b I'm not sure if you what direction you want the final velocity in the x, y, or the direction its traveling so I'll just give all 3.
Theres no change in the velocity in the x direction so its just vfx = vixcosƟ = 12cos45 = 8.49m/s
For the y direction its vfy^2 = viy^2 + 2g(Δy)
vfy = sqrt((12sin(45))^2 + 2(-9.8)(0-20m)) = 21.54m/s
The velocity the direction the stone is traveling is vf = sqrt(vx^2 + vy^2) = sqrt(8.49^2 + 21.54^2)
vf= 23.15m/s
