Question

A ball is projected horizontally from the edge of a table that is 1.15 m high, and it strikes the floor at a point 1.70 m from the base of the table.

(a) What is the initial speed of the ball?


m/s
(b) How high is the ball above the floor when its velocity vector makes a 45.0° angle with the horizontal?


m

Answer 1

Please show your work.

For part (a), Divide the 1.70 m horizontal displacement by the time required to hit the ground.

For part (b), find out where Vy = Vx = the value in part (a)

Answer 2

[AI-generated answer — Explain Bot]

To find the initial speed of the ball, we can use the fact that the horizontal component of the ball's velocity remains constant while it's in flight.

(a) To solve for the initial speed, we need to find the time it takes for the ball to hit the floor. We can utilize the equation of motion for an object thrown horizontally:

𝑥 = 𝑣₀𝑡

Where 𝑥 is the horizontal distance traveled, 𝑣₀ is the initial velocity of the ball, and 𝑡 is the time taken to hit the floor.

Given 𝑥 = 1.70 m, we need to find 𝑡. We can use the equation of motion for vertical motion:

𝑦 = 𝑣₀𝑡 + 0.5𝑔𝑡²

Where 𝑦 is the vertical distance traveled, 𝑔 is the acceleration due to gravity (-9.8 m/s²), and 𝑣₀ is the initial velocity of the ball (which is equal to zero in the vertical direction).

Since the ball is launched horizontally, 𝑦 = 1.15 m. Plugging in these values, we can solve for 𝑡:

1.15 m = 0 + 0.5(-9.8 m/s²)𝑡²

Rearranging the equation:

4.9𝑡² = 1.15 m

𝑡² = 1.15 m / 4.9 m/s²

𝑡 ≈ 0.32 seconds

Now that we know the time it takes for the ball to hit the floor, we can find the initial horizontal velocity (𝑣₀):

𝑥 = 𝑣₀𝑡

1.70 m = 𝑣₀ × 0.32 s

𝑣₀ ≈ 1.70 m / 0.32 s

So, the initial speed of the ball is approximately 5.31 m/s.

(b) To find the height of the ball when its velocity vector makes a 45.0° angle with the horizontal, we need to find the time at which this occurs. We can use trigonometry to find the time the ball spends in the air at 45.0°:

sin(45°) = vertical component of velocity / initial speed

sin(45°) = 𝑣_y / 5.31 m/s

Rearranging the equation:

𝑣_y = 5.31 m/s * sin(45°)

𝑣_y ≈ 3.76 m/s

Since the vertical component of velocity is equal to the initial vertical velocity, we have:

𝑦 = 𝑣₀𝑡 + 0.5𝑔𝑡²

1.15 m = 3.76 m/s * 𝑡 + 0.5 * (-9.8 m/s²)𝑡²

Rearranging the equation:

4.9𝑡² + 3.76𝑡 - 1.15 = 0

Solving this quadratic equation, we find two roots for 𝑡. Taking the positive root (ignoring the negative root as time cannot be negative), we get:

𝑡 ≈ 0.27 s

Now that we know the time at which the velocity vector makes a 45.0° angle with the horizontal, we can calculate the height (ℎ):

ℎ = 𝑣₀𝑡 + 0.5𝑔𝑡²

ℎ = 5.31 m/s * 0.27 s + 0.5 * (-9.8 m/s²) * (0.27 s)²

ℎ ≈ 1.20 m

Therefore, when the ball's velocity vector makes a 45.0° angle with the horizontal, its height above the floor is approximately 1.20 meters.