You can shoot an arrow straight up so that it reaches the top of a 25-m-t building. (a) How far will the arrow travel of you shoot it horizontally while pulling the bow in the same way? The arrow starts 1.45 m above the ground.(b) Where do you need to put a target that is 1.45 m above the ground in order to hit it if you aim 30° above the horizontal
Yf^2 = \Yo^2 + 2g*h = 0.
Yo^2 = -2g*h = -2*(-9.8)*25 = 490.
Yo = 22.1 m/s.
a. Yf = Yo + g*Tr = 0.
22.1 - 9.8Tr = 0.
Tr = 2.26 s. = Rise time.
Tf = Tr = 2.26 s. = Fall time.
Xo = Yo = 22.1 m/s. = Hor. velocity.
Dx = Xo*(Tr+Tf) = 22.1 * 4.52 = 100 m.
b. Range = Vo^2*sin(2A)/g.
Range = 22.1^2*sin(60)/9.8 = 43.2 m.
(a) Well, if you shoot the arrow straight up to reach the top of a 25m building, that means it will fall back down to the ground. So, if you shoot it horizontally while pulling the bow in the same way, the arrow will travel the same distance horizontally. Gravity will take care of bringing it back down vertically. It's like shooting and catching a boomerang, but without all the aerodynamics.
(b) Now, if you want to hit a target that is 1.45m above the ground, aiming 30° above the horizontal, you'll have to factor in the angle and the distance. Let's call the distance D.
To find D, we can use some trigonometry. The vertical component of the distance, Dv, is given by Dv = D * sin(30°). Since you want the target to be 1.45m above the ground, the horizontal component of the distance, Dh, will be equal to D * cos(30°).
So, the total horizontal distance, D, can be found by using Dh = D * cos(30°). Just solve for D and you'll have your answer.
But hey, remember, don't take this information too quiver-ously! It's always better to consult a professional when it comes to archery or any projectile-related activities. Safety first, folks!
(a) To determine how far the arrow will travel horizontally when shot while pulling the bow in the same way, we need to consider the horizontal velocity of the arrow.
When the arrow is shot straight up, it will experience a vertical acceleration due to gravity. However, the horizontal velocity will remain constant throughout its motion, assuming no air resistance.
We can make use of the equation of motion for horizontal motion:
distance = velocity × time
In this case, we are interested in finding the horizontal distance traveled by the arrow. Since the vertical motion of the arrow is independent of its horizontal motion, we can ignore the time of flight for now.
Given that the arrow starts 1.45 m above the ground, this initial vertical displacement does not affect the horizontal distance traveled.
Thus, we can conclude that the horizontal distance traveled by the arrow when shot horizontally is the same as if it were dropped from a height of 25 m above the ground. This is because the time of flight for both scenarios will be the same.
Using the equation for vertical motion:
distance = (1/2) × acceleration × time^2
We can solve for the time of flight by setting the displacement equal to -25 m (negative value to indicate downward motion):
-25 m = (1/2) × (-9.8 m/s^2) × time^2
Simplifying the equation:
25 m = (4.9 m/s^2) × time^2
Dividing both sides of the equation by 4.9 m/s^2:
5.1 s^2 = time^2
Taking the square root of both sides of the equation:
time = sqrt(5.1) s ≈ 2.26 s
Since the time of flight is the same for the vertical distance and the horizontal distance, we can determine the horizontal distance by multiplying the time of flight by the horizontal velocity.
Given that the horizontal component of the velocity is constant, we need to determine it.
To calculate the horizontal component of the velocity, we can use the equation:
velocity = initial velocity × cos(angle)
Since no initial velocity is mentioned, we can assume it to be nonzero given the context.
(b) To find out where to place a target that is 1.45 m above the ground to hit it when aiming 30° above the horizontal, we first need to determine the horizontal distance traveled by the arrow when it hits the target.
We can use the equation for horizontal motion:
distance = velocity × time
The horizontal distance traveled by the arrow can be calculated by multiplying the horizontal component of the velocity by the time of flight.
Given that the angle of projection is 30° above the horizontal, we can determine the horizontal component of the velocity by using trigonometry.
The horizontal component of the velocity can be calculated as:
horizontal velocity = initial velocity × cos(angle)
Since no initial velocity is mentioned, we can assume it to be nonzero given the context.
Therefore, knowing the horizontal distance traveled by the arrow and the initial vertical displacement, we can determine where to place the target.
To find the horizontal distance traveled by the arrow when shot, we can use the concept of projectile motion. When an object is shot into the air at an angle, it follows a parabolic path.
(a) To find the horizontal distance, we need to determine the time it takes for the arrow to reach the top of the 25 m building. We can use the kinematic equation for vertical displacement:
Δy = viy * t + (1/2) * a * t^2
Where:
Δy = vertical displacement (25 m)
viy = initial vertical velocity (unknown)
t = time of flight (unknown)
a = acceleration due to gravity (-9.8 m/s^2)
Since the arrow is shot straight up, the initial vertical velocity viy will be equal to the initial vertical velocity when the arrow starts from 1.45 m above the ground. We can use the kinematic equation for vertical velocity:
viy = v * sin(θ)
Where:
v = initial velocity (unknown)
θ = launch angle (unknown)
Substituting this equation into the earlier equation for Δy, we get:
25 m = (v * sin(θ)) * t + (1/2) * (-9.8 m/s^2) * t^2
Now we need to consider the horizontal motion of the arrow. The horizontal distance traveled is given by:
Δx = vix * t
Where:
Δx = horizontal distance (unknown)
vix = initial horizontal velocity (unknown)
t = time of flight (unknown)
The initial horizontal velocity vix can be found using:
vix = v * cos(θ)
Now we can substitute the known values into these equations and solve for the unknowns.
(b) To find where the target should be placed to hit it if aiming 30° above the horizontal, we need to find the horizontal distance traveled by the arrow first. Once we have that, we can find the vertical distance from the starting point to the target, which is 1.45 m above the ground.
Finding the horizontal distance is the same as in part (a). Once we have the horizontal distance, we can find the time of flight using:
t = Δx / vix
Then we can calculate the vertical distance traveled using:
Δy = viy * t + (1/2) * a * t^2
Substituting the known values, we can solve for viy:
1.45 m = (viy) * t + (1/2) * (-9.8 m/s^2) * t^2
Finally, we can use the vertical velocity to find the launch angle θ:
θ = arcsin(viy / v) - 30°
Remember to convert angles to radians when using trigonometric functions.