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Consider a ball tossed from the ground into the air at 25 m/s. At a later time, the ball is measured to be moving downward at 11 m/s. a) Find how high the ball is off of the ground at the later time. b) Find the time interval between these two events. c) Sketch graphs of position, velocity, and acceleration versus time for the ball.

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Answer:

The answer is below

Explanation:

a) According to Newton's law of motion, the distance travelled by a body going upward can be calculated using the formula:

v² = u² - 2gh

where v is the final velocity, u is the initial velocity, g is the acceleration due to gravity and h is the height of the ball.

g = 10 m/s², u = 25 m/s, v = 11 m/s

Substituting:

11² = 25² - 2(10)h

20h = 25² - 11²

20h = 504

h = 25.2 m

b) The time (t) can be gotten using:

v = u - gt

Substituting

11 = 25 - 10t

10t = 25 - 11

10t = 14

t = 1.4 s

c) the graph were plotted using an online graphing tool

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