A ball is thrown from a point 1.38 m above the ground. The initial velocity is 19.9 m/s at an angle of 33.8° above the horizontal. Find the maximum height of the ball above the ground.
Calculate the speed of the ball at the highest point in the trajectory.
looking at energy, at the highest point, the ball will have no vertical KE, so total energy will be initial PE+initialKE
topenergy=initial energy
1/2 mVhoriz^2+mg(height)=1/2 m19.9^2+mg(1.38)
where Vhoriz=19.9cosine33.8deg
solve for height.
and Vhoriz is the speed at the hightest point.
To find the maximum height of the ball above the ground, we can use the equations of projectile motion.
1. First, we need to split the initial velocity into its horizontal and vertical components.
The horizontal component (Vx) is given by Vx = V * cos(θ), where V is the initial velocity (19.9 m/s) and θ is the angle above the horizontal (33.8°).
So, Vx = 19.9 m/s * cos(33.8°). Calculating this gives us Vx ≈ 16.6 m/s.
The vertical component (Vy) is given by Vy = V * sin(θ), where V is the initial velocity (19.9 m/s) and θ is the angle above the horizontal (33.8°).
So, Vy = 19.9 m/s * sin(33.8°). Calculating this gives us Vy ≈ 11.3 m/s.
2. Next, we can use the equation to find the time it takes for the ball to reach its highest point.
The equation for vertical displacement is given by Δy = Vy * t + (1/2) * g * t^2, where Δy is the vertical displacement, Vy is the vertical component of velocity, t is the time, and g is the acceleration due to gravity (approximately 9.8 m/s^2).
Since the ball is thrown upwards, Δy is the maximum height above the ground (which we need to find), Vy is 11.3 m/s, and g is -9.8 m/s^2 (negative because it acts against the upward motion).
We need to solve this equation for t when Δy = 0, since at the highest point the vertical displacement is zero.
0 = 11.3 m/s * t + (1/2) * -9.8 m/s^2 * t^2.
This is a quadratic equation which can be solved using various methods (quadratic formula, factoring, etc.). Given the values, the quadratic equation is:
0 = -4.9 t^2 + 11.3 t.
Solving this equation gives us t = 0 s or t = 2.31 s. Since we are interested in the time it takes for the ball to reach its highest point, we choose t = 2.31 s.
3. Now that we have the time it takes for the ball to reach its highest point, we can find the maximum height above the ground.
The equation for vertical displacement is still Δy = Vy * t + (1/2) * g * t^2.
Substituting the values into the equation gives us:
Δy = 11.3 m/s * 2.31 s + (1/2) * -9.8 m/s^2 * (2.31 s)^2.
Calculating this gives us Δy ≈ 15.6 m.
Therefore, the maximum height of the ball above the ground is approximately 15.6 meters.
To calculate the speed of the ball at the highest point in the trajectory, we can use the horizontal and vertical components of velocity.
At the highest point, the vertical component of velocity (Vy) is zero, as the ball is momentarily at rest before falling downward.
The speed (magnitude of velocity) at the highest point is given by the horizontal component of velocity (Vx), which remains constant throughout the entire motion.
So, the speed of the ball at the highest point in the trajectory is approximately 16.6 m/s.