A 0.40- ball is thrown with a speed of 9.0 at an upward angle of 32 degrees. what is the speed at the highest point?
9cos32 give speed in the horizontal direction, which is what it is at the top; traveling only in the horizontal direction.
hjjh
To find the speed at the highest point of the ball's trajectory, we need to consider the vertical component of its initial velocity.
Given:
Initial velocity (v₀) = 9.0 m/s
Launch angle (θ) = 32 degrees
To determine the vertical component of the initial velocity, we can use the formula:
v₀y = v₀ * sin(θ)
where v₀y is the vertical component of the initial velocity.
Plugging in the given values:
v₀y = 9.0 m/s * sin(32°)
Using a scientific calculator or online calculator, evaluate the sine function:
v₀y ≈ 9.0 m/s * 0.5299
v₀y ≈ 4.7691 m/s
Therefore, the vertical component of the initial velocity is approximately 4.7691 m/s.
At the highest point of the projectile's trajectory, its vertical velocity becomes zero since it momentarily stops moving upwards before it starts descending. However, the horizontal component of the velocity remains constant throughout the motion.
Hence, the speed at the highest point is equal to the magnitude of the horizontal component of the initial velocity.
v_highest = v₀x
To determine the horizontal component of the initial velocity, we can use the formula:
v₀x = v₀ * cos(θ)
where v₀x is the horizontal component of the initial velocity.
Plugging in the given values:
v₀x = 9.0 m/s * cos(32°)
Using a scientific calculator or online calculator, evaluate the cosine function:
v₀x ≈ 9.0 m/s * 0.8480
v₀x ≈ 7.632 m/s
Therefore, the horizontal component of the initial velocity is approximately 7.632 m/s.
Therefore, the speed at the highest point is approximately 7.632 m/s.