An arrow is shot into the air with a speed of 40 m/s at an angle of 65 degrees.
a. Calculate the air time of the arrow
b. Calculate the maximum height the arrow reaches.
c. Calculate the range of the arrow.
To solve these problems, we can use the equations of projectile motion. Let's start by breaking down the motion of the arrow into its horizontal and vertical components.
Given:
Initial speed (v) = 40 m/s
Launch angle (θ) = 65 degrees
a. To find the air time of the arrow:
The air time (t) is the time it takes for the projectile to return to the same height from where it was launched. In kinematics, the time of flight is given by the formula:
t = 2 * (v * sin(θ)) / g
where g is the acceleration due to gravity (approximately 9.8 m/s^2).
Substituting the given values into the formula:
t = 2 * (40 * sin(65)) / 9.8
Calculating this expression will give you the air time of the arrow.
b. To find the maximum height the arrow reaches:
The maximum height (h) reached by the projectile can be calculated using the following formula:
h = (v^2 * (sin(θ))^2) / (2 * g)
Substituting the given values into the formula will give you the maximum height.
c. To find the range of the arrow:
The range (R) of the projectile is the horizontal distance covered by the arrow. It can be calculated using the formula:
R = v^2 * sin(2θ) / g
Substituting the given values into the formula will give you the range.
Please substitute the values provided and calculate the air time, maximum height, and range of the arrow.
To calculate the air time of the arrow, you can use the following formula:
T = (2 * V0 * sin(θ)) / g
Where:
T is the air time
V0 is the initial velocity of the arrow
θ is the angle of launch
g is the acceleration due to gravity (approximately 9.8 m/s²)
a. To calculate the air time, substitute the given values into the formula:
T = (2 * 40 * sin(65)) / 9.8
Calculate the numerical value of the sine of 65 degrees:
sin(65) ≈ 0.9063
Substitute the value into the formula and solve for T:
T = (2 * 40 * 0.9063) / 9.8
T ≈ 7.47 seconds
Therefore, the air time of the arrow is approximately 7.47 seconds.
b. To calculate the maximum height the arrow reaches, you can use the following formula:
H = (V0^2 * sin^2(θ)) / (2 * g)
Where:
H is the maximum height reached by the arrow
V0 is the initial velocity of the arrow
θ is the angle of launch
g is the acceleration due to gravity (approximately 9.8 m/s²)
Substitute the given values into the formula:
H = (40^2 * sin^2(65)) / (2 * 9.8)
Calculate the numerical value of the sine of 65 degrees:
sin(65) ≈ 0.9063
Substitute the values into the formula and solve for H:
H = (40^2 * 0.9063^2) / (2 * 9.8)
H ≈ 65.29 meters
Therefore, the maximum height reached by the arrow is approximately 65.29 meters.
c. To calculate the range of the arrow, you can use the following formula:
R = (V0^2 * sin(2θ)) / g
Where:
R is the range of the arrow
V0 is the initial velocity of the arrow
θ is the angle of launch
g is the acceleration due to gravity (approximately 9.8 m/s²)
Substitute the given values into the formula:
R = (40^2 * sin(2 * 65)) / 9.8
Calculate the numerical value of the sine of 2 * 65 degrees:
sin(2 * 65) ≈ 0.9063
Substitute the value into the formula and solve for R:
R = (40^2 * 0.9063) / 9.8
R ≈ 140.41 meters
Therefore, the range of the arrow is approximately 140.41 meters.
vertical component = 40 m/s * sin(65º) ... V
horizontal component = 40 m/s * cos(65º) ... H
a. flight time = 2 * V / g
b. max height = V^2 / g
c. range = H * 2 * V / g