Barky the physics dog is able to leap at a speed of 200 m/s at an angle of 35 degrees,
a) What is the maximum horizontal distance Barky can leap?
b) If Barky wanted to double his horizontal distance without changing his velocity or angle, from what height must he jump from?
initial vertical speed: 200sin35
time in air: hf=hi+vi't-4.9t^2
0=0+200sin35deg*t-4.9t^2
t= 200sin35/4.9
Horizontal distance=200cos35*t
well, if horizontal distance is doubled, t must be doubled, so
0=hi+200sin34*2t-4.9(2t)^2 solve for hi
u = 200 cos 35, forever, for parts a and b both
(that means he will have to stay in the air twice as long for b)
a) Vi = 200 sin 35
how long going up?
v = Vi - g t
v = 0 at the top
so time up Vi/g = 200 sin 35 / 9.81
total time = twice that = 400 sin 35 / 9.81
so horizontal distance = 200 cos 35 * 400 sin 35 / 9.81
b)
new time in air = twice old time = t = 800 sin 35/9.81
h = Hi + Vi t - 4.9 t^2
0 = Hi + 200 sin 35* 800 sin 35/9.81 - 4.9 (800 sin 35/9.81)^2
solve for Hi, initial height
To solve this problem, we can use the equations of projectile motion. Let's break it down step-by-step:
a) To find the maximum horizontal distance Barky can leap, we need to calculate the horizontal component of the velocity.
The horizontal component can be found using the equation:
Vx = V * cos(theta)
where Vx is the horizontal component, V is the total velocity (200 m/s), and theta is the angle (35 degrees).
Vx = 200 m/s * cos(35 degrees)
Vx ≈ 200 m/s * 0.8192
Vx ≈ 163.84 m/s
The horizontal distance can be calculated using the equation:
Horizontal distance = Vx * time
where time is the total duration of the leap.
Since we need the maximum distance, we can consider the total travel time when the projectile hits the ground again.
For a projectile launched at an angle with no air resistance, the time of flight can be calculated using the equation:
time = (2 * V * sin(theta)) / g
where g is the acceleration due to gravity (approximately 9.8 m/s^2).
time = (2 * 200 m/s * sin(35 degrees)) / 9.8 m/s^2
time ≈ (2 * 115.47 m/s) / 9.8 m/s^2
time ≈ 23.54 seconds
Horizontal distance = Vx * time
Horizontal distance = 163.84 m/s * 23.54 seconds
Horizontal distance ≈ 3861.64 meters (or approximately 3.9 kilometers)
Therefore, Barky the physics dog can leap a maximum horizontal distance of approximately 3.9 kilometers.
b) If Barky wants to double his horizontal distance without changing his velocity or angle, we can achieve this by changing the height from which he jumps.
The horizontal distance can be calculated using the equation mentioned earlier:
Horizontal distance = Vx * time
If Barky wants to double his horizontal distance, the new distance would be 2 * (163.84 m/s * 23.54 seconds).
This gives us the original horizontal distance multiplied by two.
Therefore, to find the new height from which he must jump, we need to rearrange the equation as follows:
time = (2 * V * sin(theta)) / g
Rearranging for V:
V = (time * g) / (2 * sin(theta))
To double the horizontal distance, we need:
2 * Vx * time = 2 * (163.84 m/s * 23.54 seconds)
Rearranging the equation for Vx:
Vx = (2 * (163.84 m/s * 23.54 seconds)) / time
Now we can find the new height by calculating the vertical component (Vy) using the equation:
Vy = V * sin(theta)
Vy = (163.84 m/s * 23.54 seconds) / time * sin(theta)
Vy = (163.84 m/s * 23.54 seconds) / (23.54 s) * sin(35 degrees)
Vy ≈ 3087.77 m/s
Therefore, Barky must jump from a height of approximately 3087.77 meters (or approximately 3.1 kilometers) to double his horizontal distance.
To find the maximum horizontal distance Barky can leap, we can split his initial velocity into horizontal and vertical components.
a) Maximum horizontal distance (range):
Step 1: Find the horizontal component of Barky's velocity
The horizontal component of Barky's velocity is given by Vx = V * cos(theta), where V is the magnitude of his velocity (200 m/s) and theta is the launch angle (35 degrees). Let's calculate it:
Vx = 200 m/s * cos(35 degrees) ≈ 164.3 m/s
Step 2: Calculate the time of flight
Assuming there is no air resistance, the time of flight (T) is determined by the vertical component of Barky's velocity. We can use the equation:
T = 2 * Vy / g
Since Vy is the vertical component of Barky's velocity, we need to find it first.
Vy = V * sin(theta)
Vy = 200 m/s * sin(35 degrees) ≈ 114.6 m/s
Here, g represents the acceleration due to gravity (approximately 9.8 m/s²).
T = 2 * 114.6 m/s / 9.8 m/s² ≈ 23.38 s
Step 3: Calculate the maximum horizontal distance
The maximum horizontal distance (R) is given by:
R = Vx * T
R = 164.3 m/s * 23.38 s ≈ 3836 meters
Therefore, Barky's maximum horizontal distance is approximately 3836 meters.
b) To double the horizontal distance without changing velocity or angle:
The horizontal distance is directly proportional to the time of flight (T), assuming all other variables remain constant. To double the horizontal distance, we need to double the time of flight.
2T = 2 * 23.38 s ≈ 46.76 s
Now we can use the equation for time of flight to find the required height (h) for doubling the distance:
T = 2 * Vy / g
Rearranging the equation:
h = (g * T²) / 8
Using the given value of T:
h = (9.8 m/s² * (46.76 s)²) / 8 ≈ 1106 meters
Therefore, Barky would need to jump from a height of approximately 1106 meters to double his horizontal distance.