A ball is thrown with an initial velocity of 20 meters per second at an angle of 60° with the horizontal ground surface. Find the following:
a. the horizontal velocity of the ball
b. the ball’s initial vertical velocity
c. the range of the ball
d. the maximum height of the ball
Vo = 20 m/s @ 60 Deg.
a. Xo = 20*cos60 = 10 m/s.
b. Yo = 20*sin60 = 17.3 m/s.
c. Tr = (Vf - Vo) / g,
Tr = (0 - 17.3) / -9.8 = 1.77 s. = Rise
time.
Tf = Tr = 1.77 s.
Tr + Tf = Time in flight.
R = Xo * (Tr+Tf),
R = 10 m/s. * (1.77+1.77) s. = 35.4 m.
= Range.
d. h max = (Yf^2 - Yo^2) / 2g,
h max = (0 - (17.3)^2) / -19.6=15.31 m
To find the values mentioned, we can use the following equations of motion for projectile motion:
a. The horizontal velocity of the ball remains constant throughout the motion and is equal to the initial velocity times the cosine of the angle:
Horizontal velocity = Initial velocity * cos(angle)
Horizontal velocity = 20 m/s * cos(60°)
Horizontal velocity = 20 m/s * 0.5
Horizontal velocity = 10 m/s
b. The initial vertical velocity can be found using the equation:
Vertical velocity = Initial velocity * sin(angle)
Vertical velocity = 20 m/s * sin(60°)
Vertical velocity = 20 m/s * √3/2
Vertical velocity ≈ 17.32 m/s
c. To calculate the range of the ball, we need to use the formula:
Range = (Initial velocity^2 * sin(2*angle)) / g
where g is the acceleration due to gravity (approximately 9.8 m/s^2).
Range = (20 m/s)^2 * sin(2*60°) / (9.8 m/s^2)
Range = 400 m^2/s^2 * √3 / 9.8 m/s^2
Range ≈ 22.92 m
d. The maximum height of the ball can be found using the formula:
Maximum height = (Initial velocity^2 * sin^2(angle)) / (2 * g)
Maximum height = (20 m/s)^2 * (sin(60°))^2 / (2 * 9.8 m/s^2)
Maximum height = 400 m^2/s^2 * 0.75 / 19.6 m/s^2
Maximum height ≈ 15.31 m
Therefore:
a. The horizontal velocity of the ball is 10 m/s.
b. The ball’s initial vertical velocity is approximately 17.32 m/s.
c. The range of the ball is approximately 22.92 m.
d. The maximum height of the ball is approximately 15.31 m.
To find the required values, we can use the equations of motion for projectile motion. These equations allow us to analyze the motion of an object that is launched into the air and moves along a curved trajectory.
Let's break down the problem step by step:
a. The horizontal velocity of the ball remains constant throughout its entire trajectory. It does not change due to the effect of gravity. In this case, the initial horizontal velocity is given as 20 meters per second. Therefore, the horizontal velocity of the ball is 20 meters per second.
b. The vertical velocity of the ball changes due to the effect of gravity. The initial vertical velocity can be determined by decomposing the initial velocity into its horizontal and vertical components. The vertical component of the initial velocity is given by V₀ * sin(θ), where V₀ is the magnitude of the initial velocity and θ is the launch angle. Therefore, the initial vertical velocity of the ball is 20 meters per second * sin(60°) = 17.32 meters per second.
c. To find the range of the ball, we can use the equation for the range of a projectile: R = (V₀² * sin(2θ)) / g, where R is the range, V₀ is the initial velocity, θ is the launch angle, and g is the acceleration due to gravity. Substituting the given values into the equation, we have R = (20² * sin(2 * 60°)) / 9.8. Solving this equation, we get R ≈ 41.3 meters.
d. To find the maximum height of the ball, we can use the equation for the maximum height of a projectile: H = (V₀² * sin²(θ)) / (2 * g), where H is the maximum height. Substituting the given values into the equation, we have H = (20² * sin²(60°)) / (2 * 9.8). Solving this equation, we obtain H ≈ 8.16 meters.
Thus, the answers are:
a. The horizontal velocity of the ball is 20 meters per second.
b. The ball's initial vertical velocity is 17.32 meters per second.
c. The range of the ball is approximately 41.3 meters.
d. The maximum height of the ball is approximately 8.16 meters.