You throw a beanbag in the air and catch it 2.2s later at the same place you threw it. How high did it go? What was its initial velocity?

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Physics

16 February, 01:20

You throw a beanbag in the air and catch it 2.2s later at the same place you threw it. How high did it go? What was its initial velocity?

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Cohen Giles · Answer 1

The round trip took the beanbag 2.2 s which means that moving up and down will take it only 1.1s, respectively.

a. The distance covered by the beanbag is calculated through the equation,

d = 0.5gt²

where d is distance, g is the acceleration due to gravity and t is time. Substituting the known values,

d = 0.5 (9.8) (1.1²) = 5.929 m

b. The initial velocity is calculated through the equation,

v = gt

Substituting the known values,

v = (9.8 m/s²) (1.1 s)

v = 10.78 m/s