What is the height of the tennis ball when it leaves the racket?

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A tennis ball leaves the racket horizontally at a speed of 28 m/s and travels 19.6 m before hitting the court. To determine the height from which it was struck, one must calculate the time taken to cover the horizontal distance, which is 19.6 m divided by 28 m/s. The ball falls under the influence of gravity during this time, and the height can be found using the formula for free fall. The discussion emphasizes that air resistance is typically ignored in such physics problems. The key takeaway is that the ball's horizontal speed remains constant while it falls vertically.
maiku
Kinematics in Two dimensions, from Cutnell & Johnson Physics:

A tennis ball is struck such that it leaves the racket horizontally with a speed of 28 m/s. The ball hits the court at a horizontal distance of 19.6 m from the racket. What is the height of the tenis ball when it leaves the racket?

-I've now clue how to solve this. . .

A spacecraft is traveling with a velocity of 5480 m/s along the +x direction. Two engines are turned on for a time of 842 s. One engine gives the spacecraft an acceleration in the +x direction of 1.20 m/s2, while the other gives it an acceleration in the +y direction of 8.40 m/s2. At the end of the firing, find Velocity in the x direction and the Velocity in the x direction.

-I don't even know where to begin. . .

I worked at these for an hour and a half last night, but I couldn't figure them out.
 
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Originally posted by maiku
Kinematics in Two dimensions, from Cutnell & Johnson Physics:

A tennis ball is struck such that it leaves the racket horizontally with a speed of 28 m/s. The ball hits the court at a horizontal distance of 19.6 m from the racket. What is the height of the tenis ball when it leaves the racket?

-I've now clue how to solve this. . .

I'm not a student, but I think the place to start here would be the fact that the ball falls at the same rate of acceleration no matter its horizontal speed. Since this is a textbook question, I'm geussing that wind resistance is ignored (I've always been a little jealous of that ability possesed by objects in a physics book), so the ball travels at 28 m/s throughout its path. Therefore, the time it takes the ball to travel 19.6 m at that speed is the amount of time the ball takes to fall from its initial hieght to the court surface. However far an object falls in that amount of time is the hieght from which the ball started.
 
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