an arrow leaves a bow at 30ms-1. at what two angles could the archer point the arrow if it is two reach a target at 70m away
the range of such a projectile is
R = v^2/g sin2θ
so, you need to solve for θ in
900/9.81 sin2θ = 70
To find the two possible angles at which the archer could point the arrow, we can use the equations of projectile motion.
The horizontal component of the velocity will stay constant at all points of the trajectory because there is no horizontal acceleration.
The vertical component of the velocity will change due to the acceleration due to gravity.
Let's assume the angle with the horizontal direction is θ, the initial velocity of the arrow is 30 m/s, the horizontal component of the velocity is v_x, and the vertical component of the velocity is v_y.
The horizontal distance covered by the arrow can be calculated using the equation:
Range = (v_x * time)
where time is the total time taken by the arrow to reach the target.
The time can be calculated using the equation:
time = (2 * v_y) / g
where g is the acceleration due to gravity (approximately 9.8 m/s^2).
Substituting the value of time in the first equation, we get:
Range = (2 * v_x * v_y) / g
Given that the range is 70 m and v_x = 30 m/s, we can solve for v_y:
70 = (2 * 30 * v_y) / 9.8
v_y = (70 * 9.8) / (2 * 30)
v_y = 22.68 m/s (rounded to two decimal places)
Now, to find the possible angles θ, we can use the following equations:
v_x = v * cos(θ)
v_y = v * sin(θ)
Substituting the known values, we have:
30 = 22.68 * cos(θ)
22.68 = 22.68 * sin(θ)
Solving these equations, we can find the two possible angles:
cos(θ) = 30 / 22.68 = 1.32
sin(θ) = 22.68 / 22.68 = 1.00
As the values of cos(θ) and sin(θ) cannot exceed 1, it is not possible to find two angles that satisfy these equations.
Therefore, there are no possible angles at which the archer can point the arrow to reach the target at 70 m away.
To find the two possible angles at which the archer could point the arrow, we can use the projectile motion equations. The motion of the arrow can be divided into horizontal and vertical components.
Let's assume the angle at which the archer points the arrow is θ, the initial velocity of the arrow is 30 m/s, and the distance to the target is 70 m.
1. Horizontal component:
The horizontal component of the arrow's velocity remains constant throughout its motion. Therefore, the horizontal component of the initial velocity (Vx) can be found using the equation:
Vx = V * cos(θ), where V is the initial velocity (30 m/s) and θ is the angle.
2. Vertical component:
The vertical component of the arrow's velocity changes due to the acceleration due to gravity. The equation for vertical displacement is:
Vy^2 = Vo^2 - 2g * y, where Vy is the final vertical velocity (zero when the arrow hits the target), Vo is the initial vertical velocity, g is the acceleration due to gravity (approximately 9.8 m/s^2), and y is the vertical distance traveled (70 m).
3. Finding the two angles:
Using the equations above, we can solve for θ. After finding Vx and Vy, we can find θ using the equation:
θ = arctan(Vy/Vx).
Substitute the values and solve for θ to find the two possible angles at which the archer could point the arrow.