x = xo + vot+at², v = vo + āt, v² = v² + 2ax, t = -b± √b²-4ac 2a Problem 1: A ball is thrown straight up from the ground. The ball reaches a maximum height of 5 m. For simplicity, use g = 10 m/sec². a) What is the initial speed of the ball? Answer: 10 m/sec b) What is the time it takes the ball to reach the maximum height of 5 m? Answer: 1 sec c) What is the time it takes the ball to hit the ground? Answer: 2 sec d) At what times is the ball at a height of 3 m above the ground? Answer: 0.37 sec and 1.63 sec e) What is the velocity of the ball when it is at a height of 3 m above the ground? Answer: +6.3 m/sec or -6.3 m/sec.

Principles of Physics: A Calculus-Based Text
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Author:Raymond A. Serway, John W. Jewett
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Chapter4: The Laws Of Motion
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-b+√b²-4ac
2a
x = xo + vot + ½ât², v = vô + āt, v² = v² + 2ảAx, t = ²
Problem 1:
A ball is thrown straight up from the ground. The ball reaches a maximum height of 5 m. For
simplicity, use g = 10 m/sec².
a) What is the initial speed of the ball? Answer: 10 m/sec
b) What is the time it takes the ball to reach the maximum height of 5 m? Answer: 1 sec
c) What is the time it takes the ball to hit the ground? Answer: 2 sec
d) At what times is the ball at a height of 3 m above the ground? Answer: 0.37 sec and 1.63
sec
e) What is the velocity of the ball when it is at a height of 3 m above the ground? Answer:
+6.3 m/sec or -6.3 m/sec.
Transcribed Image Text:-b+√b²-4ac 2a x = xo + vot + ½ât², v = vô + āt, v² = v² + 2ảAx, t = ² Problem 1: A ball is thrown straight up from the ground. The ball reaches a maximum height of 5 m. For simplicity, use g = 10 m/sec². a) What is the initial speed of the ball? Answer: 10 m/sec b) What is the time it takes the ball to reach the maximum height of 5 m? Answer: 1 sec c) What is the time it takes the ball to hit the ground? Answer: 2 sec d) At what times is the ball at a height of 3 m above the ground? Answer: 0.37 sec and 1.63 sec e) What is the velocity of the ball when it is at a height of 3 m above the ground? Answer: +6.3 m/sec or -6.3 m/sec.
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