A baseball is thrown horizontally off a 41-m high cliff with a speed of 18 m/s. What will be the angle in degrees between the horizontal direction and the velocity vector of the baseball just before it hits the ground? Assume that the answer will be positive and also will be the smallest angle between the horizontal and the velocity vector.
Horizontal component of velocity = 18 m/s
Vertical component of velocity, Vf, is given by
Vf²=Vi²+2*a*Δx
Vi=initial vertical velocity=0 m/s
a=acceleration due to gravity=-9.8 m/s²
Δx=-41 m
Vf=sqrt(0+2(-9.8)(-41))=28.4 m/s (downwards)
angle with horizontal
=atan(-28.4/18)
magnitude of velocity
=sqrt(-28.4²+18²)
=33.6 m/s
To find the angle between the horizontal direction and the velocity vector of the baseball just before it hits the ground, we can use trigonometry.
First, let's break down the problem and identify the given information:
- The height of the cliff (h) = 41 m
- The initial speed of the baseball (v₀) = 18 m/s
We can ignore air resistance in this problem, assuming the only force acting on the baseball is gravity.
When the baseball is thrown horizontally, its initial horizontal velocity (v₀x) is equal to the initial speed (v₀) because there is no vertical component.
The time taken for the baseball to fall from the top of the cliff to the ground can be calculated using the following equation:
h = (1/2) * g * t²
where g is the acceleration due to gravity (9.8 m/s²), and t is the time taken.
Rearranging the equation to solve for t:
t² = (2 * h) / g
t = √((2 * 41) / 9.8)
Now, we can use the time t to find the horizontal distance traveled by the baseball:
d = v₀x * t
Since the horizontal distance traveled (d) is equal to the initial horizontal velocity (v₀x) multiplied by the time (t), we can calculate the horizontal velocity using:
d = v₀x * √((2 * 41) / 9.8)
v₀x = d / √((2 * 41) / 9.8)
Now, we can calculate the angle using the tangent function:
tan(θ) = v₀y / v₀x
θ = tan^(-1)(v₀y / v₀x)
Since the baseball was thrown horizontally, the vertical component of the initial velocity (v₀y) is zero. Therefore:
θ = tan^(-1)(0 / v₀x)
θ = tan^(-1)(0) = 0 degrees
The angle between the horizontal direction and the velocity vector of the baseball just before it hits the ground is 0 degrees.