Question

An archer launches an arrow horizontally at 40 m/s from the top of a tower, striking the ground 80 meters from the base of the tower.
a) how tall is the tower?
b) what is the horizontal component of the final velocity just before landing?
c) what is the vertical component of the final velocity just before landing?
d) what is the angle of impact?

Answer 1

Initial velocity= u_x = 40 m/s. (Since it is horizontally thrown arrow), u_y = 0 m/s.

a_x = 0 m/s² , a_y = 9.8 m/s².

Also horizontal distance is given, x= 80m.

So we can find the horizontal component of final velocity-

v_fx² = v_ix² + 2*a_x * x = 40 * 40 + 2 * 0 * 80 = 1600 m/s

v_fx = 40 m/s.

We can take out time also,

x = v_ix*t + 1/2 *a_x*t²

80 = 40 * t + 1/2 * 0 * t².

t = 2 sec.

We can find the vertical component of the final velocity,

v_fy = v_iy + a_y*t = 0 + 9.8 * 2 = 19.6 m/s.

Height of the tower = v_iy * t + 1/2 * a_y* t² = 0 * 2 + 1/2 * 9.8 * 2 * 2 = 19.6 m.