An arrow is fired at an angle of 53 with a velocity of 15 m/s.
what is its range?
explain step by step please!!
and...the time is 3.06s
if it is fired from the ground, and hits the ground, then
distance=vhorizontal*timeinair
but vhorzontal is 15cos53 or 9.03m/s
time in the air is given, I think as 3.06 seconds.
Now checking the time in air given. the vertical equation is given as
vfinal=Vinitial-g *time
and vfinal is= - vinitial, so
time= 2Vinitialvertical/g
= 2*15*sin53/9.8=2.44 seconds. Hmmm. I am not certain what then you mean by time is 3.06 seconds
To find the range of the arrow, we can break down the initial velocity into its horizontal and vertical components.
Step 1: Find the horizontal component of the velocity.
Since the arrow is fired at an angle, we can find the horizontal component of the velocity using the equation:
Vx = V * cos(θ)
where Vx is the horizontal component of the velocity, V is the initial velocity, and θ is the angle of the arrow's trajectory.
In this case, the initial velocity of the arrow is 15 m/s and the angle is 53 degrees. Substituting these values into the formula, we get:
Vx = 15 m/s * cos(53°)
Vx ≈ 15 m/s * 0.6018
Vx ≈ 9.0284 m/s
Step 2: Find the time of flight.
To find the time of flight, we can use the vertical component of the velocity. The vertical component of the velocity can be found using the equation:
Vy = V * sin(θ)
where Vy is the vertical component of the velocity, V is the initial velocity, and θ is the angle of the arrow's trajectory.
Using the same values as before, we get:
Vy = 15 m/s * sin(53°)
Vy ≈ 15 m/s * 0.7961
Vy ≈ 11.9415 m/s
Since the arrow is fired into the air, it will follow a projectile motion and the time it takes to reach the maximum height will be equal to the time it takes to come back down. Therefore:
Time of flight = 2 * (Vy / g)
where g is the acceleration due to gravity, which is approximately 9.8 m/s^2.
In this case, Vy is approximately 11.9415 m/s. Substituting this value into the formula, we get:
Time of flight = 2 * (11.9415 m/s / 9.8 m/s^2)
Time of flight ≈ 2 * 1.2182 s
Time of flight ≈ 2.4364 s
Step 3: Find the range.
The range is the horizontal distance traveled by the arrow during the time of flight. To calculate the range, we multiply the horizontal component of the velocity by the time of flight:
Range = Vx * Time of flight
In this case, Vx is approximately 9.0284 m/s and the time of flight is approximately 2.4364 s. Substituting these values into the formula, we get:
Range = 9.0284 m/s * 2.4364 s
Range ≈ 21.98 m
Therefore, the range of the arrow is approximately 21.98 meters.
To find the range of the arrow, we can use the equation:
Range = (Velocity^2 * sin(2*angle)) / gravity
Here's how you can calculate it step by step:
Step 1: Convert the angle from degrees to radians
- The equation requires the angle to be in radians, so we need to convert it.
- To convert from degrees to radians, multiply the angle by π/180.
- For this problem, the angle is 53 degrees:
- Angle in radians = 53 * π/180 radians
Step 2: Calculate the sine of twice the angle
- Calculate sin(2*angle) using a scientific calculator.
- For this problem:
- sin(2*angle) = sin(2 * (53 * π/180))
Step 3: Calculate the range
- Plug the values of velocity, sin(2*angle), and gravity into the equation.
- For this problem, the velocity is 15 m/s, and assuming the acceleration due to gravity is 9.8 m/s²:
- Range = (15^2 * sin(2 * (53 * π/180))) / 9.8
Step 4: Simplify and calculate the range
- Evaluate the expression to find the range.
- Calculate the square of the velocity, then multiply it by the value obtained in step 2.
- Finally, divide the result by the acceleration due to gravity to get the range.
By following these steps, you can calculate the range of the arrow fired at an angle of 53° with a velocity of 15 m/s.