A boy 12.0 m above the ground in a tree throws a ball for his dog, who 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 m/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?
use the freefall equation to find the flight time
... h = 1/2 g t^2 + vo t + ho
... 0 = -4.90 t^2 + 0 + 12.0
(b) flight time * 8.50 m/s
(a) the dog must run as fast as the horizontal component of the ball's velocity
... 8.50 m/s
Scott pls check my question.. I need an urgent answer
To solve this problem, we can use the principles of projectile motion. Here are the steps to determine the answers:
Step 1: Calculate the time it takes for the ball to reach the ground.
Since the ball is thrown horizontally, the vertical component of its velocity is determined by the time it takes to fall from the tree to the ground. We can use the equation:
Δy = v₀y * t + (1/2) * a * t²,
where Δy is the vertical displacement, v₀y is the initial vertical velocity, t is the time, and a is the acceleration due to gravity (approximated as -9.8 m/s²).
Δy = -12.0 m (negative because the displacement is downward),
v₀y = 0 m/s (since the ball is thrown horizontally),
a = -9.8 m/s².
Plugging in these values and solving for t, we get:
-12.0 m = 0 * t + (1/2) * (-9.8 m/s²) * t².
This simplifies to:
4.9 t² = 12.0,
t² = 12.0 / 4.9,
t ≈ 1.10 s.
So, it takes approximately 1.10 seconds for the ball to reach the ground.
Step 2: Calculate the horizontal distance traveled by the ball.
Since the ball is thrown horizontally, the horizontal distance traveled is determined by the time it takes to reach the ground and the horizontal velocity of the ball.
Horizontal distance = horizontal velocity * time,
Horizontal distance = 8.50 m/s * 1.10 s,
Horizontal distance ≈ 9.35 m.
So, the ball will land approximately 9.35 meters away from the tree.
Step 3: Calculate the speed of the dog.
To catch the ball just as it reaches the ground, the dog needs to cover the horizontal distance faster than the ball. Since the time for the ball to reach the ground is approximately 1.10 seconds, the dog needs to cover the horizontal distance in less than 1.10 seconds.
Speed = Distance / Time,
Speed = 9.35 m / 1.10 s,
Speed ≈ 8.50 m/s.
So, the dog must run at approximately 8.50 m/s to catch the ball just as it reaches the ground.
To summarize:
(a) The dog must run at approximately 8.50 m/s.
(b) The dog will catch the ball approximately 9.35 meters away from the tree.
To solve this problem, we can use the principles of kinematics and consider the horizontal and vertical motions separately.
Let's start with part (b) of the question: finding the distance from the tree where the dog catches the ball. Since the boy throws the ball horizontally, the horizontal motion of the ball is unaffected by gravity. Therefore, we can ignore the horizontal motion and focus on the vertical motion only.
We know that the initial vertical position of the ball is 12.0 m above the ground, the acceleration due to gravity is 9.8 m/s^2 (assuming Earth's gravity), and the final vertical position is 0 m (on the ground). We need to find the time it takes for the ball to reach the ground.
Using the formula for vertical displacement in free fall:
Δy = v0*t + (1/2)*a*t^2
Where:
Δy = final vertical position - initial vertical position = 0 - 12.0 = -12.0 m (negative because it is downward)
v0 = initial vertical velocity = 0 (since the boy throws it horizontally)
a = acceleration due to gravity = -9.8 m/s^2 (negative because it is downward)
t = time
Plugging in the values:
-12.0 = 0*t + (1/2)*(-9.8)*t^2
Simplifying and rearranging the equation:
4.9*t^2 = 12.0
Dividing both sides by 4.9:
t^2 = 12.0 / 4.9
t^2 = 2.45
Taking the square root of both sides:
t ≈ √2.45 ≈ 1.56 s
So, it takes approximately 1.56 seconds for the ball to reach the ground.
Now, let's move on to part (a) of the question: finding the speed at which the dog must run to catch the ball just as it reaches the ground. Since the dog starts running the instant the ball is thrown, we can assume that the time taken by the ball to reach the ground is the same as the time taken by the dog to reach the ball.
The horizontal distance traveled by the ball is given by:
d = v0*t
Where:
d = horizontal distance
v0 = horizontal velocity = 8.50 m/s (since the boy throws the ball horizontally)
t = time taken for the ball to reach the ground ≈ 1.56 s (from our previous calculation)
Plugging in the values:
d = 8.50 * 1.56 ≈ 13.26 m
So, the distance from the tree where the dog catches the ball is approximately 13.26 meters.
To find the speed at which the dog must run, we can use the formula for speed:
Speed = Distance / Time
Speed = 13.26 m / 1.56 s ≈ 8.49 m/s
Therefore, the dog must run at a speed of approximately 8.49 m/s to catch the ball just as it reaches the ground.