apa harapan anda mengenai pembelajaran PJOK selama 1 tahun ke depan

Time in the air of the marble using the equation y=1/2gt^2, re-arranged to make t the subject - t = √(2y/g). Therefore… t = √(0.94x2/9.81) = 0.4377… where g=9.81.
Using the equation x=Ut (the horizontal displacement) 2.35=u x 0.4377.. re-arrange to make u = x/t. Therefore.. u=2.35/0.4377 = 5.368m/s =5.37 = 3 significant figures

Answered by jackstattershal | 2024-06-10

The initial velocity of the marble as it leaves the launching device is 5.35 m/s ​ . ;

Answered by avantikar | 2024-06-12

The initial velocity of the marble as it leaves the launching device is approximately 5.36 m/s. This was determined by calculating the time it takes to fall from 94.0 cm and then using that time to find the horizontal speed. The marble travels 2.35 m horizontally during its fall.
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Answered by avantikar | 2024-09-30

jawaban:bisa lebih baik daripada tahun yang lalu,dan bisa cara ubah penampilan di kelas menjadi lebih baik dari pada tahun yang lalu, dan bisa lebih ceria dan penuh semangat dan motivasi bagi para mahasiswa,.

Answered by gresiasaragih98 | 2025-07-23