Question

A stone is catapulted at time t = 0, with an initial velocity of magnitude 18.6 m/s and at an angle of 37.0° above the horizontal. What are the magnitudes of the (a) horizontal and (b) vertical components of its displacement from the catapult site at t = 1.01 s? Repeat for the (c) horizontal and (d) vertical components at t = 1.76 s, and for the (e) horizontal and (f) vertical components at t = 5.11 s. Assume that the catapult is positioned on a plain horizontal ground.

Answer 1

a)

consider the motion along the horizontal direction

V_ox = initial velocity in horizontal direction = 18.6 Cos37

t = time = 1.76 sec

X = horizontal component of displacement

X = V_ox t = (18.6 Cos37 ) (1.76) = 26.14 m

b)

consider the motion along the vertical direction

V_oy = initial velocity in vertical direction = 18.6 Sin37

t = time = 1.76 sec

a = - 9.8

Y = vertical component of displacement

Y= V_oy t + (0.5) a t² = (18.6 Sin37 ) (1.76) + (0.5) (-9.8) (1.76)² = 4.52 m

c)

consider the motion along the horizontal direction

V_ox = initial velocity in horizontal direction = 18.6 Cos37

t = time = 1.01 sec

X = horizontal component of displacement

X = V_ox t = (18.6 Cos37 ) (1.01) = 15 m

d)

consider the motion along the vertical direction

V_oy = initial velocity in vertical direction = 18.6 Sin37

t = time = 1.01 sec

a = - 9.8

Y = vertical component of displacement

Y= V_oy t + (0.5) a t² = (18.6 Sin37 ) (1.01) + (0.5) (-9.8) (1.01)² = 6.31 m