Question

A stone is catapulted at time t = 0, with an initial velocity of magnitude 18.6 m/s and at an angle of 41.6° above the horizontal. What are the magnitudes of the (a) horizontal and (b) vertical components of its displacement from the catapult site at t = 1.06 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 = 5.10 s. Assume that the catapult is positioned on a plain horizontal ground.

Answer 1

Vo = 18.6m/s[41.6o]
Xo = 18.6*Cos41.6 = 13.91 m/s
Yo = 18.6*sin41.6 = 12.35 m/s

a. X = Xo=13.91 m/s and does not change.
Dx = Xo*t = 13.91 * 1.06 = 14.74 m.

b. Y = Yo + g*t = 12.35 - 9.8*1.06 = 1.962 m/s
h = Yo*t + 0.5g*t^2
Yo=12.35 m/s, t = 1.06 s, g=-9.8 m/s^2.
Solve for h.

c. Same as part "a" except t = 1.76 s.

d. Same as part b except t = 1.76 s.

e. Same as part a except t = 5.10 s.

f. Same as part b except t = 5.10 s.

Answer 2

Note: The Y component of the velocity
decreases with time and is zero at the
maximum height. The velocity increases
as the stone falls.

c. Dx = Xo*t = 13.91 * 1.76 = 24.48 m.

d. Y = Yo + g*Tr = 0
12.35 - 9.8Tr = 0
9.8Tr = 12.35
Tr = 1.26 s. = Rise time or time to reach max ht.

h max = Yo*Tr + 0.5g*t^2 =
12.35*1.26 - 4.9*1.26^2 = 7.78 m.

Tf = 1.76-1.26 = 0.50 s. = Fall time.

h = ho - 0.5g*t^2 = 7.78 - 4.9*0.5^2 =
6.56 m.

e. The stone hits gnd before 5.10 s:
T = 2Tr = 2*1.26 = 2.52 s. = Time in air

Dx = Xo*T = 13.91 * 2.52 = 35.1 m. = Max
range.

f. The stone hits gnd in 2.52 s. So its'
ht. at 5.10 s = 0