A floating ice block is pushed through a displacement d = (13.5 m)i + (-14.3 m)j along a straight embankment by rushing water, w
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Answer:
The initial velocity vector of the ball is;
= 25·i + 19.6·j
Explanation:
The given parameters are;
The time of flight of the ball = 4 seconds
The horizontal distance from the plate at which the ball is caught = 100 m
Let 'u', represent the initial velocity, we have;
u × cos(θ) × t = u × cos(θ) × 4 s = 100 m
u × cos(θ) = 25 m/s...(1)
∴ 2·u·sin(θ) = 9.8 m/s² × 4 s = 39.2 m/s
tan(θ) = 1.568/2 = 0.784
θ = arctan(0.784) ≈ 38.096°
The direction of the velocity vector of the ball, θ ≈ 38.096°
From equation (1), we have;
u × cos(θ) = 25 m/s
∴ u = 25 m/s/cos(θ) = 25 m/s/cos(arctan(0.784)) = 31.7672787629 m/s
The magnitude of the initial velocity vector, u = 31.7672787629 m/s
The vertical component of the initial velocity, = u × sin(θ)
∴ = 31.7672787629 × sin(arctan(0.784))
Therefore, the initial velocity vector of the ball is approximately, v = 31.767 m/s in a direction 38.096° above the horizontal, from which we have;
u = uₓ·i + ·j = 25·i + 19.6·j
= 25·i + 19.6·j