A person who is 1.1 m tall throws a ball with a speed of 3.5 m/s at an angle of 52 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 = 3.5m/s @ 52o
X = 3.5*cos52 = 2.15 m/s.
Y = 3.5*sin52 = 2.76 m/s.
wrong
To find the x and y components of the velocity of the ball before it strikes the ground, we can use the following equations:
Vx = Vinitial * cos(θ)
Vy = Vinitial * sin(θ) - (g * t)
where:
Vx is the x-component of the velocity
Vy is the y-component of the velocity
Vinitial is the initial velocity of the ball
θ is the angle of projection
g is the acceleration due to gravity (10 m/s^2)
t is the time it takes for the ball to strike the ground
First, let's find the time it takes for the ball to reach the ground. Since we know the initial vertical velocity and the acceleration due to gravity, we can solve for time using the equation:
0 = Vinitial * sin(θ) - (g * t)
Rearranging the equation, we get:
t = Vinitial * sin(θ) / g
Now, let's calculate the time:
t = (3.5 m/s) * sin(52°) / (10 m/s^2)
t ≈ 0.915 s
Using this value of time, we can now find the x and y components of the velocity:
Vx = Vinitial * cos(θ)
Vx = (3.5 m/s) * cos(52°)
Vx ≈ 2.258 m/s
Vy = Vinitial * sin(θ) - (g * t)
Vy = (3.5 m/s) * sin(52°) - (10 m/s^2 * 0.915 s)
Vy ≈ 1.872 m/s
Therefore, the x-component of the velocity of the ball is approximately 2.258 m/s, and the y-component of the velocity is approximately 1.872 m/s.