Vertical problem:
Vi = 49 sin 30 = 24.5 m/s
v = Vi - 9.81 t
at top v = 0
0 = 24.5 - 9.81 t
t = 2.50 seconds upward
so time in air = 2*2.5 = 5 seconds
h = Vi t - 4.9 t^2
at t 2.5
h = 24.5*2.5 - 4.9(6.25)
= 61.25 - 30.625
= 30.625 meters high
Horizontal:
u = 49 cos 30 forever or
at least until the ground hits it.
so
range = 49 cos 30 * 5 seconds
an arrow is shot at a 30 degree angle with the horizontal. it has a velocity of 49 m/s how high will the arrow go? what horizontal distance will the arrow travel? and how long will the arrow be in the air?
3 answers
Vo = 49m/s[30o] = Initial velocity.
Xo = 49*Cos30 = 42.4 m/s = Ver. component.
Yo = 49*sin30 = 24.5 m/s = Hor. component.
A. Y^2 = Yo^2 + 2g*h.
0 = 24.5^2 + (-19.6)*h,
h = 30.63 meters.
B. Range = Vo^2*sin(2A)/g
Range = 49^2*sin(60)/9.8 = 212.2 m.
C. Y = Yo + g*Tr.
0 = 24.5 + (-9.8)Tr,
Tr = 2.5 s. = Rise time.
Tf = Tr = 2.5 s = Fall time.
Tr+Tf = 2.5 + 2.5 = 5.0 s. = Time in air.
Xo = 49*Cos30 = 42.4 m/s = Ver. component.
Yo = 49*sin30 = 24.5 m/s = Hor. component.
A. Y^2 = Yo^2 + 2g*h.
0 = 24.5^2 + (-19.6)*h,
h = 30.63 meters.
B. Range = Vo^2*sin(2A)/g
Range = 49^2*sin(60)/9.8 = 212.2 m.
C. Y = Yo + g*Tr.
0 = 24.5 + (-9.8)Tr,
Tr = 2.5 s. = Rise time.
Tf = Tr = 2.5 s = Fall time.
Tr+Tf = 2.5 + 2.5 = 5.0 s. = Time in air.
the answer was good