A baseball is thrown with a speed of 40m/s at an angle of 30 degrees. Neglecting
friction what is the speed of the ball in m/s when it reaches a height of 10 meters?
Vy = sqrt(Uy^2 - 2gh); Uy = U sin(theta)
Vx = Ux = U cos(theta)
So |V| = sqrt(Vy^2 + Vx^2) = sqrt(Uy^2 - 2gh + Ux^2) = sqrt(U^2 - 2gh) = sqrt(40^2 - 2*9.8*10) = 37.46998799 = 37.47 ANS B.
Note that U^2 = Uy^2 + Ux^2.
To find the speed of the ball when it reaches a height of 10 meters, we can use the principles of projectile motion. We'll split the motion into horizontal and vertical components.
Step 1: Find the initial vertical velocity (Vy).
The initial velocity (V) of the ball can be split into horizontal (Vx) and vertical (Vy) components using trigonometry.
Vy = V * sin(theta)
Vy = 40 m/s * sin(30 degrees)
Vy = 20 m/s
Step 2: Find the time taken (t) to reach a height of 10 meters.
We can use the equation:
y = Vy * t - 1/2 * g * t^2
where y is the vertical displacement (10 m) and g is the acceleration due to gravity (9.8 m/s^2).
Rearranging the equation, we get:
t^2 - (2Vy/g) * t + 2y/g = 0
Substituting the values:
t^2 - (2 * 20 m/s / 9.8 m/s^2) * t + 2 * 10 m / 9.8 m/s^2 = 0
Solving this quadratic equation, we find:
t ≈ 2.04 seconds
Step 3: Find the final velocity (Vf) when the ball reaches a height of 10 meters.
The final velocity at a certain height can be calculated using the equation:
Vf = Vy - g * t
Substituting the values:
Vf = 20 m/s - 9.8 m/s^2 * 2.04 s
Vf ≈ 0 m/s
Therefore, the speed of the ball when it reaches a height of 10 meters is approximately 0 m/s, meaning it momentarily comes to a stop at its peak height.
To determine the speed of the ball when it reaches a height of 10 meters, we need to break down the motion of the ball into horizontal and vertical components.
Given:
Initial speed of the ball (v₀) = 40 m/s
Launch angle (θ) = 30 degrees
Height reached by the ball (h) = 10 meters
Acceleration due to gravity (g) = 9.8 m/s² (assuming no air resistance)
1. Calculate the time taken for the ball to reach the height of 10 meters:
To find the time (t) taken to reach the maximum height, we can use the equation of motion in the vertical direction:
h = v₀ * sin(θ) * t - (1/2) * g * t²
Rearranging the equation, we get:
10 = 40 * sin(30) * t - (1/2) * 9.8 * t²
5t² - 20t + 19.6 = 0 (dividing both sides by 2)
Solving this quadratic equation, we find that t ≈ 2.195 seconds.
2. Calculate the vertical component of the velocity (v_y) at the maximum height:
Using the equation of motion v = u + gt, where u is the initial vertical velocity:
v_y = v₀ * sin(θ) - g * t
v_y = 40 * sin(30) - 9.8 * 2.195
v_y ≈ 18.53 m/s
3. Calculate the final velocity when the ball reaches 10 meters:
At the highest point, the final vertical velocity becomes zero. The horizontal component of the velocity remains constant throughout the motion.
The final velocity (v_f) can be found using the Pythagorean theorem:
v_f = sqrt((v_x)² + (v_y)²)
v_f = sqrt((v₀ * cos(θ))² + (v_y)²)
v_f = sqrt((40 * cos(30))² + (18.53)²)
v_f ≈ 46.96 m/s
Therefore, the speed of the ball when it reaches a height of 10 meters (neglecting friction) is approximately 46.96 m/s.