A person who is 1.1 m tall throws a ball with a speed of 5.2 m/s at an angle of 56 degrees above the horizontal. What are the x and y components of the velocity (in m/s) of the ball the instant before it strikes the ground? The acceleration due to gravity is 10 m/s2
Vo = 5.2m/s[56o].
Xo = 5.2*Cos56 = 2.91 m/s.
Yo = 5.2*sin56 = 4.31 m/s.
h=-Yo^2/2g + 1.1=-(4.31)^2/-20 + 1.1
= 2.03 m. Above gnd.
Y^2 = Yo^2 + 2g*h = 0 + 20*2.03 = 40.6
Y = 6.37 m/s.
X = Xo = 2.91 m/s.
To find the x and y components of the velocity of the ball, we can use trigonometry and break down the initial velocity into horizontal and vertical components.
The initial velocity (v) of the ball can be broken down into its x-component (vx) and y-component (vy) as follows:
vx = v * cos(θ)
vy = v * sin(θ)
Where:
v = initial velocity of the ball (5.2 m/s)
θ = angle of projection above the horizontal (56 degrees)
Plugging in the values, we have:
vx = 5.2 m/s * cos(56 degrees)
vy = 5.2 m/s * sin(56 degrees)
Now we can calculate these values:
vx = 5.2 m/s * cos(56 degrees) ≈ 2.682 m/s
vy = 5.2 m/s * sin(56 degrees) ≈ 4.200 m/s
Therefore, the x-component of the velocity is approximately 2.682 m/s, and the y-component of the velocity is approximately 4.200 m/s.