3] A ball is thrown from the top of a building with an initial velocity of 20.0 m/s straight upward, at an initial height of 50.0 m above the ground. The ball just misses the edge of the roof on its way down, as shown in Figure 2.20. Determine (a) the time needed for the ball to reach its maximum height, (b) the maximum height, (c) the time needed for the ball to return to the height from which it was thrown and the velocity of the ball at that instant, (d) the time needed for the ball to reach the ground, and (e) the velocity and position of the ball at t5 5.00 s. Neglect air drag.(please see the figure 2.20 in the book)

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3] A ball is thrown from the top of a building with an initial velocity of 20.0 m/s straight upward,
at an initial height of 50.0 m above the ground. The ball just misses the edge of the roof on its
way down, as shown in Figure 2.20. Determine (a) the time needed for the ball to reach its
maximum height, (b) the maximum height, (c) the time needed for the ball to return to the
height from which it was thrown and the velocity of the ball at that instant, (d) the time needed
for the ball to reach the ground, and (e) the velocity and position of the ball att 55.00 s. Neglect
air drag.(please see the figure 2.20 in the book)
Transcribed Image Text:3] A ball is thrown from the top of a building with an initial velocity of 20.0 m/s straight upward, at an initial height of 50.0 m above the ground. The ball just misses the edge of the roof on its way down, as shown in Figure 2.20. Determine (a) the time needed for the ball to reach its maximum height, (b) the maximum height, (c) the time needed for the ball to return to the height from which it was thrown and the velocity of the ball at that instant, (d) the time needed for the ball to reach the ground, and (e) the velocity and position of the ball att 55.00 s. Neglect air drag.(please see the figure 2.20 in the book)
t = 2.04 s
20.4 m
Imax =
v = 0
t = 0
Yo =0
Vo = 20.0 m/s
t = 4.08 s
y =D0
v = -20.0 m/s
t = 5.00 s
y = -22.5 m
v = -29.0 m/s
50.0 m
t = 5.83 s
y = -50.0 m
v = -37.1 m/s
Transcribed Image Text:t = 2.04 s 20.4 m Imax = v = 0 t = 0 Yo =0 Vo = 20.0 m/s t = 4.08 s y =D0 v = -20.0 m/s t = 5.00 s y = -22.5 m v = -29.0 m/s 50.0 m t = 5.83 s y = -50.0 m v = -37.1 m/s
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