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?
vertical problem
it falls from 12 m up, no initial velocity
v = -g t
h = 12 - 4.9 t^2
so at sea surface
t = sqrt (12/4.9) and v = - 9.81 t
now horizontal problem
constant u = 50
time t we just got above
so
speed = sqrt (50^2+v^2)
tan angle = v/u
Thanks Damon
How do you calculate V^2 ?
v = -g t
v = -g t
if the first equation in the vertical problem, falling object
To determine the speed and angle of the arrow when it enters the sea, we can use the principles of projectile motion.
Step 1: Determine the time it takes for the arrow to reach the sea.
Since the arrow is fired horizontally, its initial vertical velocity is zero. The vertical distance the arrow travels is equal to the height of the cliff, which is 12 meters. We can use the equation of motion in the vertical direction:
y = u * t + (1/2) * a * t^2
where:
y = vertical displacement (12 m)
u = initial vertical velocity (0 m/s)
a = acceleration due to gravity (-9.8 m/s^2)
t = time
Plugging in the known values:
12 = 0 * t + (1/2) * (-9.8) * t^2
12 = -4.9 * t^2
Solving for t, we get:
t^2 = 12 / 4.9
t ≈ 1.23 seconds
Step 2: Determine the horizontal distance covered by the arrow.
The horizontal distance covered is given by the equation:
x = u * t
where:
x = horizontal displacement (distance covered, which we want to find)
u = initial horizontal velocity (50 m/s)
t = time (1.23 s)
Plugging in the values:
x = 50 * 1.23
x ≈ 61.5 meters
Step 3: Determine the speed of the arrow when it enters the sea.
The speed of the arrow when it enters the sea is the magnitude of its velocity vector, which is the square root of the sum of its horizontal and vertical velocities squared.
The horizontal velocity remains constant throughout the motion, so it will be 50 m/s.
The vertical velocity is affected by gravity and can be calculated using the equation:
v_y = u * t + a * t
where:
v_y = vertical velocity at time t
u = initial vertical velocity (0 m/s)
t = time (1.23 s)
a = acceleration due to gravity (-9.8 m/s^2)
Plugging in the values:
v_y = 0 * 1.23 + (-9.8) * 1.23
v_y ≈ -12 m/s
Now we can calculate the speed:
speed = √(v_x^2 + v_y^2)
= √(50^2 + (-12)^2)
≈ √(2500 + 144)
≈ √(2644)
≈ 51.4 m/s
Step 4: Determine the angle to the horizontal when the arrow enters the sea.
The angle to the horizontal can be calculated using the equation:
θ = tan^(-1)(v_y / v_x)
θ = tan^(-1)(-12 / 50)
≈ tan^(-1)(-0.24)
≈ -13.04 degrees
Note: The negative sign indicates that the arrow is entering the sea at an angle below the horizontal.
Therefore, the speed of the arrow when it enters the sea is approximately 51.4 m/s, and the angle to the horizontal is approximately -13.04 degrees.