Question

A baseball is thrown with an initial velocity of 100 m/s at an angle 30 above the horizontal How far from the throwing point will it attain at іtѕ original level?​

Answer 1

30 Above the horizontel far from the throwing point is 130m/s

Answer 2

Answer:The attempt at a solution: vix = 100 cos 30° => 86.6 m/s viy = 100 cos 30° => 50 m/s To find the distance we need to use v = d t v=dt = d = vt Now in order to find the time we can use; y = v i y t + 1 2 a y t 2 y=viyt+12ayt2 y = 0 y=0 since the ball is coming back to its original height. = 0 = ( 50 m / s ) . t + 1 2 ( − 9.8 m / s 2 ) t 2 0=(50m/s).t+12(−9.8m/s2)t2 Am I right.....? I can't exactly see how to obtain the time out of this… Thanks

Explanation: hope this helps u plz mark me as the brainliest