An arrow is fired horizontally at a speed of 50m/s from the top of a vertical cliff overlooking the sea. The height of the cliff is 12m. Determine
The speed and angle to the horizontal of the arrow when it enters the sea?
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To determine the speed and angle to the horizontal when the arrow enters the sea, we can break down the problem into two parts: horizontal motion and vertical motion.
First, let's consider the horizontal motion. Since the arrow is fired horizontally, there are no horizontal forces acting on it. Therefore, the horizontal velocity of the arrow remains constant throughout its flight. In this case, the initial horizontal velocity of the arrow is given as 50 m/s.
Next, let's analyze the vertical motion of the arrow. We know that the arrow is fired from a height of 12 m above the sea. The only force acting vertically on the arrow is the force due to gravity, which causes the arrow to accelerate downwards.
Using the kinematic equation for vertical motion:
h = ut + (1/2)gt^2
Where:
h is the vertical displacement (12 m),
u is the initial vertical velocity (unknown),
g is the acceleration due to gravity (9.8 m/s^2),
and t is the time taken for the arrow to reach the sea (unknown).
Since the arrow is fired horizontally, there is no initial vertical velocity (u) because the arrow is not initially moving vertically. Therefore, we can rewrite the equation as:
h = (1/2)gt^2
Plugging in the given values:
12 = (1/2)(9.8)t^2
Solving for t:
t^2 = (2 * 12) / 9.8
t^2 = 2.4489
t ≈ 1.564 s
Now, we can determine the initial vertical velocity (u) using the equation:
v = u + gt
Since the arrow has reached the sea, the final vertical velocity (v) is zero when entering the water. Therefore:
0 = u + (9.8)(1.564)
Solving for u:
u = - (9.8)(1.564)
u = -15.254 m/s (negative sign indicates the direction is downward)
Now that we have the initial vertical velocity, we can use it to find the overall speed and angle to the horizontal when the arrow enters the sea.
The overall speed (v) of the arrow is the vector sum of its horizontal and vertical velocities. In this case, the horizontal velocity remains constant at 50 m/s, and the vertical velocity u = -15.254 m/s:
v = sqrt((horizontal velocity)^2 + (vertical velocity)^2)
v = sqrt((50)^2 + (-15.254)^2)
v ≈ 52.47 m/s
Finally, the angle (θ) to the horizontal when the arrow enters the sea can be calculated using trigonometry:
θ = atan2(vertical velocity, horizontal velocity)
θ = atan2(-15.254, 50)
θ ≈ -16.79 degrees (negative sign indicates downward direction)
Therefore, the speed of the arrow when it enters the sea is approximately 52.47 m/s, and the angle to the horizontal is approximately -16.79 degrees (downward).