Physics, asked by supidebcparthi, 1 year ago

The position of a projectile launched from the origin at t=0 is given by r= 40 i^+50j^ at t=2secr→=(40i^+50t=2s. If the projectile was launched at an angle thetaθfrom the horizontal , then thetaθis

Answers

Answered by jaspreetsinghhhh
61
2ux  = 40 => 4x = 20

50 = 24y - 1/2 * 10 * 22 => 4y = 35

tan theta = uy / ux = 35 / 20 = 7 / 4

theta = tan -1 [ 7/4 ]
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