Question

Leonardo launches at a height of 156 [cm] with his rubber band, a rubber band so that it acquires a speed of 86.431 [ft / s] with an angle of 36 °. If Harold, he goes directly to Leonardo at a speed of 12.35 [km / h] in the same plane as the trajectory of the rubber band: a) How far should Harold from Leonardo be to capture the rubber band with his mouth, if Harold's height to the mouth is 3.46528 [yards]. b) with what magnitude of speed the rubber band enters the mouth of Harold.
solve with vectors i, j, k.

Answer 1

h1 = 156cm = 1.56m
h2 = 3.46528 yd = 3.46528x0.9144 = 3.17m

s = h2-h1 = 1.61 j m

u = 86.431 ft/s @36 deg = 26.34 m/s @ 36 deg

= 26.34cos36 i + 26.34sin36 j

= 21.31 i + 15.48 j m/s

a = -9.81 j m/s^2

using second kinematical equation in vertical direction

s = ut + 1/2 at^2

1.61 = 15.48 t - 4.905 x t^2
t = 3.05 sec

Using speed = dist/time in horizontal direction

21.31 = x1 / 3.05

x1 = 65 i m

u2 = 12.35 km/h = 12.35 x 5/18 = 3.43 m/s

x2 = u2t = 3.43 x 3.05 = 10.46 i m

So, the total distance that Harold be away from Leonardo is

x = x1 + x2 = 65 + 10.46

= 75.46 m