Answer to Question #79044 in Mechanics | Relativity for Murenzi Frank

Question

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.

Answer 1

x=20 cos⁡〖45 〗 t,y=20 sin⁡〖45 〗 t-1/2 gt^2
y/x=tan⁡30
So,
(20 sin⁡〖45 〗 t-1/2 gt^2)/(20 cos⁡〖45 〗 t)=tan⁡30
(20 sin⁡〖45 〗-1/2(9.8)t)/(20 cos⁡〖45 〗 )=tan⁡30
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.