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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?

Respuesta :

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