A stone is catapulted at time t = 0, with an initial velocity of magnitude 21.1 m/s and at an angle of 39.9° above the horizontal. What are the magnitudes of the (a) horizontal and (b) vertical components of its displacement from the catapult site at t = 1.08 s? Repeat for the (c) horizontal and (d) vertical components at t = 1.75 s, and for the (e) horizontal and (f) vertical components at t = 5.09 s. Assume that the catapult is positioned on a plain horizontal ground.

(a) X = 17.48 m

(b) Y = 8.903 m

(c) X = 28.33 m

(d) Y = 8.68 m

(e) X = 44.68 m

(f) Y = 0 m

Explanation:

Given;

magnitude of initial velocity, v = 21.1 m/s

angle of projection, θ = 39.9°

the horizontal component of the velocity,

the vertical component of the velocity,

(a)  the horizontal components of its displacement;

gravitational influence on horizontal direction is negligible, g = 0

(b) the vertical components of its displacement;

(c) the horizontal components of its displacement;

(d) the vertical components of its displacement;

(e)  the horizontal components of its displacement;

(f) the vertical components of its displacement;

this displacement is not possible because it is beyond zero vertical level; it shows that the stone will hit the ground before 5.09 s.

When the stone hits the ground Y = 0 m

At zero vertical displacement (Y = 0 m), the time at that position will be calculated as;

The horizontal displacement at this time is given by;

X = vₓt

X = (16.187 x 2.76)

X = 44.68 m

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