Rock physics problem

Question

A rock is thrown from the top of a tower with a speed of 2.5 x 10¹ m/s and at a downward angle of 30.0⁰ to the horizontal, as shown, as shown. What are the horizontal and vertical components of the ball’s initial velocity?

Answer 1

I do not see the picture, but hope could easily imagine it. First of all I would like to notice that in any case of motion we characterize it with vector quantities like velocity, displacement, acceleration. As any vector, they have magnitude and direction. We need in a right way find their magnitudes.

To do all problems involving vectors we usually select coordinate axes to go from vector to scalar expression of the variables.

To find the component of the vector, we use the formula a_x = a cos Θ, where a - magnitude of the vector and Θ is the angle between the directions of vector a and axis.

In our particular case we will direct x-axis to the right (assuming that the body after we through it moves in this direction) and y-axis vertically down.

The vector of initial velocity v₀ will have 2 components: v_0x and v_oy.

At any given moment the change of velocity with time in case of free fall could be described as

v(t) = v₀ + gt,

where v - instantaneous velocity,

v₀ - initial velocity,

g - acceleration due to gravity.

Yes, if we through the body horizontally, or at any arbitrary angle, release it from rest, throw it vertically up or down, the body is in free fall. Because the only force that acts on the body - is force of gravity.

For x-axis we can write v_x = v_0x + g_x

In the same manner for y-axis v_y = v_0y + g_y

Vector v₀ forms with x-axis angle Θ = 30º. Hence,

its component on x-axis

v_0x = v₀ cos Θ,

And on y-axis

v_0y = v₀ sin Θ

Plugging in the given values, we obtain

v_0x = 2.5 cos 30º = 2.2 m/s

v_0y = 2.5 sin 30º = 1.25 m/s

Answer: 2,2,m/s and 1.25 m/s

My addition for teachers, tutors, educators and students!

Now let's find components of vector g of acceleration due to gravity. We see that

g_x = 0

It is understandable, because the angle between vector g and x-axis is 90°, and cos of 90° is 0.

g_y = g

This is also not a big surprise, because angle between vector g and y-axis is 0, and cos of 0° is 1.

Please, please stop telling that the acceleration of free fall g is negative. It is wrong!

The vector g of free fall all the time has direction down! Its magnitude is 9.81 m/s². Magnitude cannot be negative!

Like any vector, it could have the component with negative, positive, or zero value. It all depends on selection of the direction of the axis and angle between this axis and vector g.

Finally, we can write that components of vector of velocity

v_x = v₀

v_{y =} gt

It means that the horizontal component of initial velocity is constant, it is not changing with time.

Vertical component is linearly depends on time.

And at any given time the magnitude of instantaneous velocity

v = √ ( v_x² + v_y²)