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Chapter 3: Problem 49

A boy \(12.0 \mathrm{~m}\) above the ground in a tree throws a ball for his dog. which is standing right below the tree and starts running the instant the ball is thrown. If the boy throws the ball horizontally at \(8.50 \mathrm{~m} / \mathrm{s}\) (a) how fast must the dog run to catch the ball just as it reaches the ground, and (b) how far from the tree will the dog catch the ball?

Short Answer

Expert verified
(a) The dog must run at 8.50 m/s; (b) The dog will catch the ball 13.26 m from the tree.

Step by step solution

01

Identify Given Data

The initial height of the ball is 12.0 m. The ball is thrown horizontally with a speed of 8.5 m/s. We need to calculate how fast and how far the dog must run to catch the ball when it reaches the ground.
02

Calculate Time of Flight

Using the formula for the time of flight for a falling object: \[ t = \sqrt{\frac{2h}{g}} \] where \( h = 12 \text{ m} \) and \( g = 9.81 \text{ m/s}^2 \). Substitute to find \( t \):\[ t = \sqrt{\frac{2(12)}{9.81}} \approx 1.56 \text{ seconds} \] .
03

Calculate Horizontal Distance

The horizontal distance (\( x \)) covered by the ball is given by:\[ x = v_{x} \, t \] where \( v_{x} = 8.5 \text{ m/s} \) and \( t = 1.56 \text{ s} \). Substitute to find \( x \):\[ x = 8.5 \times 1.56 \approx 13.26 \text{ m} \].
04

Determine Dog's Speed

Since the dog must reach the ball when it lands, and the horizontal distance to cover is 13.26 m within 1.56 seconds, the dog's speed is:\[ v_{\text{dog}} = \frac{x}{t} \] Substitute to find \( v_{\text{dog}} \): \[ v_{\text{dog}} = \frac{13.26}{1.56} \approx 8.50 \text{ m/s} \].

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Horizontal Motion
When a projectile, like the ball in this exercise, is thrown with an initial horizontal speed, it moves parallel to the ground without any acceleration in that direction. This occurs because, barring air resistance, there are no forces acting in the horizontal plane once it has been launched. In this situation, the horizontal velocity of the ball remains constant. Hence, for our ball:
  • Initial horizontal speed: 8.5 m/s
  • Horizontal speed remains the same throughout the motion
This constant speed results in the ball covering ground steadily over time, and it factors into predicting how far the ball will cast a shadow on the ground before landing.
Time of Flight
The time of flight is how long the ball stays in the air before hitting the ground. In projectile motion, only the vertical component affects this time, as the ball's horizontal velocity has no impact on how long it takes to fall. To find the time of flight:- We used the formula \[t = \sqrt{\frac{2h}{g}}\]
  • where \( h \) is the initial height, 12 m in this case
  • \( g \) is the acceleration due to gravity, 9.81 m/s虏
Using these values, the time taken for the ball to reach the ground is approximately 1.56 seconds. This is critical because it tells us how much time the dog has to reach the ball.
Vertical Motion
Vertical motion in this context advances differently from horizontal motion due to gravity's influence. The ball is initially at rest in the vertical direction and accelerates downward. - The initial vertical velocity is 0 m/s because the ball was thrown horizontally. - Gravity acts to accelerate the ball at 9.81 m/s虏. This accelerating force causes the ball to drop downwards, determining how fast it can move downwards and for how long it stays airborne. At the end of this period, which we calculated as 1.56 seconds, the ball complete its fall to the ground from the 12.0 m height.
Dog's Speed Calculation
To figure out how fast the dog needs to run, the key is to relate the horizontal motion of the ball to the distance the dog must cover. We already know:- The horizontal distance the ball travels is 13.26 m.- The time to cover this distance is 1.56 seconds.Using these, the speed of the dog can be computed using the simple formula\[v_{\text{dog}} = \frac{x}{t}\]where
  • \(x\) is the horizontal distance covered
  • \(t\) is the time during which this distance is covered
Thus, substituting gives us a required speed of about 8.50 m/s. This speed is exactly how fast the dog needs to run to ensure catching the ball precisely when it lands.

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Most popular questions from this chapter

A firefighting crew uses a water cannon that shoots water at \(25.0 \mathrm{~m} / \mathrm{s}\) at a fixed angle of \(53.0^{\circ}\) above the horizontal. The firefighters want to direct the water at a blaze that is \(10.0 \mathrm{~m}\) above ground level. How far from the building should they position their cannon? There are \(t w o\) possibilities; can you get them both? (Hint: Start with a sketch showing the trajectory of the water.)

At an air show, a jet plane has velocity components \(v_{x}=625 \mathrm{~km} / \mathrm{h}\) and \(v_{y}=415 \mathrm{~km} / \mathrm{h}\) at time \(3.85 \mathrm{~s}\) and \(v_{x}=838 \mathrm{~km} / \mathrm{h}\) and \(v_{y}=\) \(365 \mathrm{~km} / \mathrm{h}\) at time \(6.52 \mathrm{~s}\). For this time interval, find (a) the \(x\) and \(y\) components of the plane's average acceleration and (b) the magnitude and direction of its average acceleration.

A tennis player hits a ball at ground level, giving it an initial velocity of \(24 \mathrm{~m} / \mathrm{s}\) at \(57^{\circ}\) above the horizontal. (a) What are the horizontal and vertical components of the ball's initial velocity? (b) How high above the ground does the ball go? (c) How long does it take the ball to reach its maximum height? (d) What are the ball's velocity and acceleration at its highest point? (e) For how long a time is the ball in the air? (f) When this ball lands on the court, how far is it from the place where it was hit?

A batted baseball leaves the bat at an angle of \(30.0^{\circ}\) above the horizontal and is caught by an outfielder \(375 \mathrm{ft}\) from home plate at the same height from which it left the bat. (a) What was the initial speed of the ball? (b) How high does the ball rise above the point where it struck the bat?

A football is kicked from ground level at a speed of \(25 \mathrm{~m} / \mathrm{s}\). If it reaches a maximum height of \(24 \mathrm{~m},\) at what angle was it kicked (relative to horizontal)?
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