A particle is projected with a velocity of 20 m/s at an angle of 45 to the horizontal. Find its range on a plane inclined at 30 to the horizontal when projected upwards.
1
Expert's answer
2018-07-12T10:44:08-0400
x=20 cos〖45 〗 t,y=20 sin〖45 〗 t-1/2 gt^2 y/x=tan30 So, (20 sin〖45 〗 t-1/2 gt^2)/(20 cos〖45 〗 t)=tan30 (20 sin〖45 〗-1/2(9.8)t)/(20 cos〖45 〗 )=tan30 Thus, t=1.22 s. The range on a plane inclined at 30 to the horizontal: R=√((20 sin〖45 〗 t-1/2 gt^2 )^2+(20 cos〖45 〗 t)^2 ) R=√((20 sin〖45 〗 (1.22)-1/2 (9.8) (1.22)^2 )^2+(20 cos〖45 〗 (1.22))^2 )=20 m.
"assignmentexpert.com" is professional group of people in Math subjects! They did assignments in very high level of mathematical modelling in the best quality. Thanks a lot
Comments
Leave a comment