fired at angle T and speed s
u = s cos T
Vi = s sin T
range = 45 meters
Hi = 1.39
vertical problem:
v = Vi - 9.81 t
h = 1.39 + Vi t - 4.9 t^2
0 = 1.39 + Vi t - 4.9 t^2
horizontal problem:
u = s cos T forever
so
45 = u cos T * t
t = 45 / cos T
putting together:
0 = 1.39 + s sin T * t - 4.9 t^2
0 = 1.39 + s sin T * 45/cos T - 4.9 (45/cos T)^2
Now to continue you need another point. Like how high is A ???
An archer releases an arrow from a shoulder height of 1.39 m. When
the arrow hits the target 18 m away, it hits point A. When the target is
removed, the arrow lands 45 m away. Find the maximum height of the
arrow along its parabolic path.
2 answers
I don't know, it's not given