Question

A stone is thrown with an initial speed of 15 m/s at an angle of 53° above the horizontal from the top of a 40-m building. If g = 9.8 m/s2 and air resistance is negligible, then what is the magnitude of the horizontal component of velocity as the rock strikes the ground?

Answer 1

[AI-generated answer — Bot GPT 3.5]

To find the magnitude of the horizontal component of velocity as the rock strikes the ground, we first need to find the time it takes for the rock to hit the ground.

Let's use the following equations of motion to solve this problem:

v = u + at
s = ut + 0.5at^2
v^2 = u^2 + 2as

Where:
v = final velocity
u = initial velocity
a = acceleration
t = time
s = displacement

First, let's find the time it takes for the rock to hit the ground.

The vertical component of the initial velocity can be calculated as:
v_y = u sin(theta)
v_y = 15 sin(53)
v_y ≈ 11.64 m/s

Next, let's find the time it takes for the rock to hit the ground, using the equation:
s = ut + 0.5at^2

Substitute the known values:
-40 = 11.64 * t - 0.5 * 9.8 * t^2

Rearrange the equation to form a quadratic equation:
-40 = 11.64t - 4.9t^2
4.9t^2 - 11.64t - 40 = 0

Solving the quadratic equation, we find two possible solutions for t:
t = 4.42 sec (ignoring the negative solution)

Now, let's find the horizontal component of the final velocity using the equation:
v_x = u cos(theta)

Substitute the known values:
v_x = 15 cos(53)
v_x ≈ 9.26 m/s

Therefore, the magnitude of the horizontal component of velocity as the stone strikes the ground is approximately 9.26 m/s.

Answer 2

correct answer, but you did not have to go through all those vertical component calculations, since the horizontal component is just a constant

Answer 3

[AI-generated answer — Bot GPT 3.5]

You are correct. Since acceleration in the horizontal direction is 0 and there are no horizontal forces acting on the stone (neglecting air resistance), the horizontal component of velocity remains constant throughout the motion.

Therefore, the magnitude of the horizontal component of velocity as the stone strikes the ground is the same as the initial horizontal component, which is given by:
v_x = u cos(theta)

Substitute the known values:
v_x = 15 cos(53)
v_x ≈ 9.26 m/s

Thus, the magnitude of the horizontal component of velocity as the stone strikes the ground is approximately 9.26 m/s. Thank you for pointing out that the detailed vertical component calculations were unnecessary in this case.