Respuesta :
The trajectory (parabola) for a "particle" of water follows from:
h(t) = v(0) sin(angle) t - 1/2 g t^2
x(t) = v(0) cos(angle) t
From the second equation t = x / (v(0) cos(angle) )
and substitution in the first to eliminate t gives the parabola h(x):
h(x) = x tan(angle) - (g/(2v(0)^2 cos(angle)^2) x^2
h(x) = 0 is solved by x=0 (where the trajectory starts) and at
x = 2 v(0)^2 tan(angle) cos^2(angle) / g
x = 2 v(0)^2 sin(angle) cos(angle) / g
x = v(0)^2 sin(2 angle) / g [[ using sin(2 ) = 2 sin(s) cos(s) ]]
For the first question (the 2.0m distance question) you have therefore to solve for angle the equation
2.0= 2*(6.5)^2 sin(2 angle) / 9.81
sin(2 angle) = 0.23
angle = 1/2 * arcsin(0.23) = 6.65 degrees
I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!
h(t) = v(0) sin(angle) t - 1/2 g t^2
x(t) = v(0) cos(angle) t
From the second equation t = x / (v(0) cos(angle) )
and substitution in the first to eliminate t gives the parabola h(x):
h(x) = x tan(angle) - (g/(2v(0)^2 cos(angle)^2) x^2
h(x) = 0 is solved by x=0 (where the trajectory starts) and at
x = 2 v(0)^2 tan(angle) cos^2(angle) / g
x = 2 v(0)^2 sin(angle) cos(angle) / g
x = v(0)^2 sin(2 angle) / g [[ using sin(2 ) = 2 sin(s) cos(s) ]]
For the first question (the 2.0m distance question) you have therefore to solve for angle the equation
2.0= 2*(6.5)^2 sin(2 angle) / 9.81
sin(2 angle) = 0.23
angle = 1/2 * arcsin(0.23) = 6.65 degrees
I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!
