A gray kangaroo can bound across level ground with each jump carrying it 8.4m from the takeoff point. Typically the kangaroo leaves the ground at a 23 angle.
1. Provide knowns and unknown values.
2. What is the initial velocity or takeoff speed?
3. What is its maximum height above the ground?
Please provide formula and numerical answer.
1. Known values:
- Distance covered in each jump: 8.4m
- Angle at which the kangaroo leaves the ground: 23°
Unknown values:
- Initial velocity or takeoff speed
- Maximum height above the ground
2. To find the initial velocity or takeoff speed, we can use the following formula:
v₀ = (d / t) * cos(θ)
Where:
- v₀ is the initial velocity or takeoff speed
- d is the distance covered in each jump (8.4m)
- t is the time spent in the air in each jump
- θ is the angle at which the kangaroo leaves the ground (23°)
However, we don't have the time spent in the air in each jump, so we need to find it first.
Using the vertical motion formula:
h = (v₀ * t) - (0.5 * g * t²)
Where:
- h is the maximum height above the ground
- v₀ is the initial velocity or takeoff speed
- t is the time spent in the air in each jump
- g is the acceleration due to gravity (approximately 9.8 m/s²)
3. To find the maximum height above the ground, we need to determine the time spent in the air in each jump and then substitute it into the above formula.
Unfortunately, the information provided does not allow us to find the time spent in the air. Therefore, it is not possible to calculate the initial velocity or takeoff speed and the maximum height above the ground. Sorry about that!
1. Known values:
- Distance traveled in each jump: 8.4 m
- Angle of takeoff: 23 degrees
Unknown values:
- Initial velocity or takeoff speed (Vi)
- Maximum height above the ground (Hmax)
2. To find the initial velocity, we can use the range formula:
Range (R) = (Vi^2 * sin(2 * θ)) / g,
where:
- Vi is the initial velocity or takeoff speed
- θ is the angle of takeoff
- g is the acceleration due to gravity (approximately 9.8 m/s^2)
Substituting the given values:
8.4 m = (Vi^2 * sin(2 * 23))/ 9.8
Simplifying the equation:
Vi^2 * sin(2 * 23) = 8.4 * 9.8
Vi^2 * sin(46) = 82.32
Using the sine double-angle identity:
Vi^2 * 2 * sin(23) * cos(23) = 82.32
Simplifying further:
Vi^2 = 82.32 / (2 * sin(23) * cos(23))
Taking the square root of both sides:
Vi = √(82.32 / (2 * sin(23) * cos(23)))
Calculating the value of Vi using a scientific calculator or math software:
Vi ≈ 11.32 m/s
Therefore, the initial velocity or takeoff speed of the kangaroo is approximately 11.32 m/s.
3. To find the maximum height above the ground, we can use the following formula:
Hmax = (Vi^2 * sin^2(θ)) / (2 * g),
where:
- Vi is the initial velocity or takeoff speed
- θ is the angle of takeoff
- g is the acceleration due to gravity (approximately 9.8 m/s^2)
Substituting the given values:
Hmax = (11.32^2 * sin^2(23)) / (2 * 9.8)
Calculating Hmax using a scientific calculator or math software:
Hmax ≈ 1.85 m
Therefore, the maximum height above the ground reached by the kangaroo is approximately 1.85 m.
1. Known values:
- Distance covered in each jump (d) = 8.4 m
- Launch angle (θ) = 23°
- Acceleration due to gravity (g) = 9.8 m/s²
Unknown values:
- Initial velocity or takeoff speed (v₀)
- Maximum height above the ground (h)
2. To find the initial velocity or takeoff speed (v₀), we can use the following formula:
v₀ = d / (cosθ * t)
In this case, the time (t) is the total time spent in the air during each jump.
To find the time (t), we can use the formula:
d = v₀ * cosθ * t
Rearranging this formula gives us:
t = d / (v₀ * cosθ)
Now we can substitute this expression for time back into the initial velocity formula:
v₀ = d / (cosθ * (d / (v₀ * cosθ)))
Simplifying this equation gives us:
v₀ = d * v₀ * cosθ / d, or
1 = v₀ * cosθ / 1
Simplifying further gives us:
v₀ = 1 / cosθ
Substituting the value of the launch angle (θ = 23°) into the equation, we get:
v₀ = 1 / cos(23°)
v₀ = 1 / 0.921
Calculating this gives us:
v₀ ≈ 1.085 m/s
So, the initial velocity or takeoff speed is approximately 1.085 m/s.
3. To find the maximum height above the ground (h), we can use the following formula:
h = (v₀ * sinθ)² / (2 * g)
Substituting the values into the equation gives us:
h = (1.085 * sin(23°))² / (2 * 9.8)
Calculating this gives us:
h ≈ 0.085 m
Therefore, the maximum height above the ground is approximately 0.085 m.
say speed is S
then horizontal velocity u = S cos 23
Vi, initial vertical velocity = S sin 23
t is total time in air and t/2 is time to max height
d = distance = u t
so
8.4 = u t = .920 S t
S t = 9.13
at top
time = t/2 = 9.13/2S = 4.57/S
v = 0 = Vi - 9.8 (4.57/S)
0 = .391 S - 44.8/S
.391 S^2 = 44.8
S = 10.7 m/s takeoff speed
h = Vi (t/2) - 4.9 (t/2)^2
t/2 = 4.57/S = .427
Vi = .391 (10.7) = 4.18
h = 4.18(.427) - 4.9 (.427)^2
h = .891