Question

A ball is thrown from a field with a speed of 12 m/s at an angle of 45 degree with the horizontal. At what distance will it hit the field again? Take g= 10 m/s².

Answer 1

Voy=12×root of 2÷2

Voy=6×root of 2

0=6×(root of 2)–10t

tymax=6×(root of 2)÷10 seconds

txmax=2tymax

txmax=2×6×(root of 2)÷10

txmax=12×(root of 2)÷10

Vox= 12×root of 2÷2

Vx=6×root of 2

Distance= v×t

Distance=6×(root of 2)×12×(root of 2)÷10

Distance=72×2÷10

Distance=14.4 meters

It will hit the field again after 14.4 meters

Answer 2

hlw mate ur ans ..
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