Question

In the figure below Thabo initially at rest throws a ball upward at an angle 𝜃𝑜 with an

initial speed 𝑣𝑜. He tries to catch up to the ball by speeding-up at a constant
acceleration 𝑎 for a time interval ∆𝑡1 and then continues to run at a constant speed for
a time interval ∆𝑡2. He catches the ball at exactly the same height he threw the ball. Let
𝑔 be the gravitational constant. What was Thabo’s acceleration 𝑎?

Answer 1

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Answer 2

[AI-generated answer — Explain Bot]

To find Thabo's acceleration 𝑎, we can analyze the motion of the ball he throws. Let's break down the motion into three phases:

Phase 1: Throwing the ball upward
During this phase, the ball is thrown with an initial speed 𝑣₀ at an angle 𝜃₀. The acceleration due to gravity, 𝑔, acts downwards. Since the ball reaches the same height it was thrown from, it means that the time taken to reach the maximum height is equal to the time taken to return to that height.

Phase 2: Accelerating to catch up
Thabo tries to catch up to the ball by speeding up at a constant acceleration 𝑎 for a time interval ∆𝑡₁. Since the ball is in projectile motion and already changing its velocity due to gravity, Thabo's acceleration will depend on the initial velocity and angle of projection.

Phase 3: Constant speed
After accelerating, Thabo continues to run at a constant speed for a time interval ∆𝑡₂. Thabo's speed during this phase will be constant, and hence his acceleration will be zero.

Given this information, we need to determine Thabo's acceleration 𝑎 in Phase 2. To do that, we can use the following equations of motion for the vertical component of the ball's motion:

1. Vertical displacement (Δ𝑦) = 0 (since the ball returns to the same height it was thrown from)
2. Vertical initial velocity (𝑣₀sin𝜃₀)
3. Vertical final velocity at maximum height (𝑣ᵣ = 0 at the highest point of the ball's motion due to the change in direction)
4. Time taken to reach maximum height (𝑡ₘ)

Applying the equation for vertical displacement, we have:

Δ𝑦 = 𝑣₀𝑡ₘsin𝜃₀ - (1/2)𝑔𝑡ₘ²

Since Δ𝑦 = 0, we can rearrange the equation to solve for time:

(1/2)𝑔𝑡ₘ² = 𝑣₀𝑡ₘsin𝜃₀ [Equation 1]

Next, we need to find Thabo's acceleration 𝑎 in terms of the initial velocity and angle of projection. Since acceleration is the rate of change of velocity, we can use the equation:

𝑎 = ∆𝑣/∆𝑡₁

During Phase 2, Thabo tries to catch up to the ball, so his vertical displacement will be the same as the ball's (Δ𝑦 = 0). Therefore, we can relate Thabo's acceleration and time interval (∆𝑡₁) to the ball's initial velocity and angle:

0 = 𝑣₀𝑡ₘsin𝜃₀ + 1/2 𝑎 (∆𝑡₁)² [Equation 2]

We now have two equations (Equation 1 and Equation 2) with two unknowns (𝑡ₘ and 𝑎). By solving these equations simultaneously, we can find Thabo's acceleration 𝑎.

Answer 3

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