Question

A ball is at an altitude of 130m on a tall, thin radio tower of uniform cross-sectional area and it is shot from a paintball gun at a velocity of 10 m/s down toward the ground at an angle 10 degrees from the vertical. How long is the ball in the air after leaving the muzzle of the paintball gun before it strikes the ground below? How fast is the ball moving when it strikes the ground? How far horizontally from the base of the tower does the ball strike the ground? Draw a diagram. Draw and label an axis. Show your work. Your answer should have the correct units. You should setup the problem, draw axes, list all of your equations and known values, and then solve the problem.

Answer 1

The ball is in the air for approximately 2.54 seconds. It strikes the ground directly below the tower and has a velocity of approximately 25.1 m/s.

Step-by-step explanation:

To solve this problem, we need to break down the motion of the ball into its vertical and horizontal components. Let's first consider the vertical motion. The ball is launched at an initial vertical velocity of 10 m/s, and it will accelerate downwards due to gravity. The ball's final vertical position is 130 m below its starting position, so we can use the equation:

Δy = v0yt + (1/2)gt2

where Δy is the displacement in the vertical direction, v0y is the initial vertical velocity, t is the time, and g is the acceleration due to gravity (-9.8 m/s2). Solving for t gives:

t = (√(2Δy/g)) - (√(2h/g))

where h is the height of the tower. Plugging in the values, we find that t ≈ 2.54 s. This is the total time the ball is in the air.

Next, let's consider the horizontal motion. The ball is launched with an initial horizontal velocity of 0 m/s, so it will maintain a constant horizontal velocity of 0 m/s throughout its motion. Therefore, the horizontal distance travelled by the ball is simply the horizontal component of the initial velocity times the time:

Δx = v0xt

where Δx is the horizontal displacement, v0x is the initial horizontal velocity, and t is the time. Plugging in the values, we find that Δx = 0 m, which means that the ball strikes the ground directly below the tower.

Lastly, let's consider the velocity of the ball when it strikes the ground. We can use the fact that the vertical velocity of the ball when it lands is given by:

vy = v0y + gt

Plugging in the values, we find that vy ≈ -25.1 m/s (negative sign indicates a downward direction). The velocity of the ball just before it hits the ground can be found using the Pythagorean theorem:

v = √(vx2 + vy2)

Plugging in the values, we find that v ≈ 25.1 m/s. Therefore, the ball is moving at approximately 25.1 m/s when it strikes the ground.