Final answer:
At its highest point, the 0.20 kg ball will have a horizontal velocity of 5.74 m/s. The ball reaches a maximum height of 0.82 m.
Step-by-step explanation:
The student asked about the speed of the ball at its highest point and how high it goes when thrown with an initial speed of 7.0 m/s at a 35-degree angle upward. The speed at the highest point is solely the horizontal component because the vertical component of the velocity becomes zero at that point.
To find the horizontal velocity, we use the cosine component of the initial velocity: horizontal velocity = initial velocity × cos(theta), which is 7.0 m/s × cos(35°) = 5.74 m/s. This is the speed of the ball at its highest point.
To determine the maximum height, we can use the vertical component and the formula for free-fall motion under gravity. The initial vertical velocity is 7.0 m/s × sin(35°) = 4.01 m/s. The maximum height h is calculated using the formula h = (initial vertical velocity)² / (2 × gravitational acceleration), which gives us h = (4.01 m/s)² / (2 × 9.8 m/s²) = 0.82 m.