An archer shoots an arrow with a velocity of 49.5 m/s at an angle of 52.0¡Æ with the horizontal. An assistant standing on the level ground 150 m downrange from the launch point throws an apple straight up with the minimum initial speed necessary to meet the path of the arrow.
(a) What is the initial speed of the apple?
(b) At what time after the arrow launch should the apple be thrown so that the arrow hits the apple?
Physics Quick! - Henry, Monday, August 20, 2012 at 7:32pm
a. Vo = 49.5 m/s @ 52o.
Xo = 49.5*cos52 = 30.5 m/s.
Yo = 49.5*sin52 = 39.0 m/s.
hmax = (Y^2-Yo^2)/2g.
hmax = (0-1521) / -19.6 = 77.6 m.
Tr = (Y-Yo)/g.
Tr = (0-39 / -9.8 = 3.98 s. = Rise time.
Xo*t = Dx.
t = Dx / Xo = 150 / 30.5 = 4.92 s.=Time
to reach 150 m range.
h = ho - 0.5g*t^2
h = 77.6 - 4.9*(4.92-3.98)^2 = 73.3 m.=
Dist. from arrow to gnd @ 150 m range.
Vo^2 + 2g*h = V^2.
Vo^2 - 19.6*73.3 = 0
Vo^2 - 1436.68
Vo=37.9 m/s.=Initial velocity of apple.
b. Tr = (V-Vo)/g.
Tr = (0-37.9) / -9.8 = 3.87 s. = Rise
time of apple.
T = 4.92 - 3.87 = 1.05 s after arrow is launched