horizontal speed is constant from takeoff to crash
u= 1.5 /0.547
if the total initial speed is s
then that u you just calculated is S cos 45
and Vi, initial speed up = S sin 45
but
sin 45 = cos 45
so
Vi = u
as time goes on of course, v = Vi - g t
but that was not part of the question
I don't understand how to answer this problem or even which equation to insert it into:
"A Dancer leaps across the stage in a beautiful grand jete. The ballerina leaps at an angle of 45° and is able to jump a horizontal distance of 1.5m in 0.547s. What is her initial x and initial y velocity?"
3 answers
Range = Vo^2*sin(2A)/g = 1.5 m.
Vo = Initial velocity.
A = 45o.
g = 9.8m/s^2.
Xo = Vo*Cos 45.
Yo = Vo*sin 45.
Vo = Initial velocity.
A = 45o.
g = 9.8m/s^2.
Xo = Vo*Cos 45.
Yo = Vo*sin 45.
i need help with the same problem except mine is 25 degrees not 45