一小球做平抛运动的初速度为2m/s,落地速度与水平方向的夹角为37°,求小球抛出到落地点的距离和所用的时间

题目

一小球做平抛运动的初速度为2m/s,落地速度与水平方向的夹角为37°,求小球抛出到落地点的距离和所用的时间
sin37°＝0.6,cos37°＝0.8,g＝10m/s2

答案

velocity in horizontal is constant 2m/s
velocity at the end is 2m/s / cos37° = 2.5m/s
velocity in vertical at the end is 2.5m/s * sin37° = 1.5m/s
velocity in vertical at the beginning is 0m/s
from 0m/s to 1.5m/s,it costs 1.5 / g = 0.15 s (the time)
horizontal distance is constant velocity 2m/s * 0.15s = 0.3m
vertical distance is 1/2 * g * 0.15 * 0.15 = 0.1125 m
distance from the begining to the end is 'square root' of (0.3*0.3 + 0.1125*0.1125) = 0.32m (the distance)