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Lucas throws a football with an initial upward velocity component of 16.0 m/s and a horizontal velocity component of 20.0 m/s. Ignore air resistance, a) how much time is required for the football to reach the highest point of the trajectory? b) How high is this point? c) How much time (after being thrown) is required for the football to return to its original level? d) How far has it travelled horizontally during this time?

Respuesta :

Answer:

The answers to the question are

a) 1.632 s

b) 13.05 m

c) 3.265 s

d) 65.31 m.

Explanation:

To solve the question we note that

v = u - g·t

Where u = initial velocity = 16 m/s

v = final velocity = 0 m/s

t = time

Therefore 0 = 16 - 9.8×t or t = 16/9.8 = 1.632 s

b) We have

v² = u² - 2·g·h

Where h = height

Therefore 0 = 16² -2×9.81×h

or h = 16²/19.62 = 13.05 m

c) The time required for the ball to return to its original level is

2 × time to maximum height = 2×1.632 = 3.265 s

d) Distance traveled horizontally is

Horizontal Velocity  =  20 m/s

Distance =Horizontal Velocity × Time = 20×3.265 = 65.31 m