A person is riding on a flatcar traveling at a constant speed of 200 km/s (Fig.). He wishes to throw a ball through a stationary hoop 90 km above the height of his hands in such a manner that the ball will move horizontally as it passes through the hoop. He throws the ball with a speed of 250 m/s with respect to himself.
(a) How many time required after he releases the ball will it pass through the hoop?
(b) What must the vertical component of the initial velocity of the ball?
(c) At what horizontal distance in front of the hoop must he release the ball?
(d) When the ball leaves the man's hands, what is the direction of its velocity relative to the
frame of reference of the flatcar? Relative to the frame of reference of an observer standing
on the ground?
No flatcar goes 200 km/second. Maybe a rocket.
Surely you mean 9 meters up ???? I can hardly throw a single mile.
what is initial vertical speed component Vi and v, the vertical speed during trrip?
v = Vi - g t
at the top horizontal means v = 0
(1/2)m Vi^2 = m g h
Vi^2 = 2 g h = 2 * 9.81 * 9 = 176.58
Vi = 13.3 m/s
Now how long ? our t?
v = 0 = Vi - 9.81 t
t = 13.3 / 9.81 = 1.35 seconds rise time
Oh come on. He did not throw at 250 meters/s
I give up, too many typos to figure out.
To answer these questions, we first need to analyze the motion of the ball relative to the flatcar and then convert the velocities and distances to a common frame of reference.
Let's start with part (a), which asks for the time required for the ball to pass through the hoop.
(a) The ball will follow a parabolic trajectory because it has both a vertical component and horizontal component of motion. The time it takes for the ball to reach its peak height and fall back down to the hoop is the time required. To find this time, we can use the vertical distance traveled by the ball.
The vertical distance traveled by the ball is the height of the hoop, which is 90 km. Let's convert that to meters for consistency. 90 km = 90,000 m.
Assuming there is no vertical acceleration and neglecting air resistance, we can use the equation for vertical displacement of an object in projectile motion:
Δy = Vyi * t - (1/2) * g * t^2
Here, Δy is the vertical displacement, Vyi is the initial vertical velocity of the ball, t is the time, and g is the acceleration due to gravity (approximately 9.8 m/s^2).
We need to solve for t when Δy = 90,000 m.
90,000 = Vyi * t - (1/2) * 9.8 * t^2
Now, let's move on to part (b).
(b) To determine the vertical component of the ball's initial velocity, we can use the horizontal distance it travels. Since the ball moves horizontally as it passes through the hoop, the horizontal distance traveled by the ball will be equal to the distance traveled by the flatcar during the time the ball is in the air.
To find this distance, we can use the equation for distance:
d = V * t
Here, d is the horizontal distance, V is the horizontal velocity of the flatcar, and t is the time we calculated in part (a).
The horizontal velocity of the flatcar is given as 200 km/s, but we need to convert it to m/s for consistency. 200 km/s = 200,000 m/s.
Now, let's solve for d.
d = 200,000 * t
Now, let's move on to part (c).
(c) The horizontal distance in front of the hoop where the ball must be released can be found by considering the time it takes for the ball to reach the hoop and the horizontal distance it travels during that time.
From part (a), we determined the time it takes for the ball to pass through the hoop. Let's call this time t.
To find the horizontal distance traveled by the ball, we can again use the equation for distance:
d = Vx * t
Here, d is the horizontal distance, Vx is the horizontal component of the ball's initial velocity, and t is the time we calculated in part (a).
Since the ball moves horizontally as it passes through the hoop, the horizontal distance traveled by the ball will be equal to the distance traveled by the flatcar during the time the ball is in the air. Therefore, we can use the same horizontal velocity of the flatcar, 200,000 m/s, for Vx.
Now, let's solve for d.
d = 200,000 * t
Finally, let's move on to part (d).
(d) When the ball leaves the man's hands, its velocity relative to the frame of reference of the flatcar is entirely horizontal since it moves horizontally through the hoop. Therefore, its direction of velocity relative to the flatcar is purely horizontal.
Relative to the frame of reference of an observer standing on the ground, the ball's velocity will have both horizontal and vertical components. As the ball moves through the hoop, its direction of velocity relative to the ground will depend on the angle at which it was thrown. However, since the question states that the ball is thrown with a speed of 250 m/s with respect to the person, we can assume that the ball is thrown horizontally relative to the ground. Therefore, the direction of velocity relative to the ground is purely horizontal.
To summarize:
(a) We need to solve for the time required for the ball to pass through the hoop using the vertical distance traveled by the ball.
(b) The vertical component of the initial velocity of the ball is not directly given and can be determined using the horizontal distance traveled by the ball (equal to the distance traveled by the flatcar).
(c) The horizontal distance in front of the hoop where the ball must be released is equal to the distance traveled by the flatcar during the time the ball is in the air.
(d) The ball's direction of velocity relative to the flatcar is purely horizontal, and relative to an observer on the ground, it is also purely horizontal assuming the ball is thrown horizontally.